5 - Structural Dynamics

Structural Analysis IV Chapter 5 – Structural DynamicsDr. C. Caprani 1 Chapter 5 - Structural Dynamics 5.1 Introduction ......................................................................................................... 3 5.1.1 Outline of Structural Dynamics ..................................................................... 3 5.1.2 An Initial Numerical Example ....................................................................... 5 5.1.3 Case Study – Aberfeldy Footbridge, Scotland .............................................. 8 5.1.4 Structural Damping ...................................................................................... 10 5.2 Single Degree-of-Freedom Systems ................................................................. 11 5.2.1 Fundamental Equation of Motion ................................................................ 11 5.2.2 Free Vibration of Undamped Structures...................................................... 16 5.2.3 Computer Implementation & Examples ...................................................... 20 5.2.4 Free Vibration of Damped Structures .......................................................... 26 5.2.5 Computer Implementation & Examples ...................................................... 30 5.2.6 Estimating Damping in Structures ............................................................... 33 5.2.7 Response of an SDOF System Subject to Harmonic Force ........................ 35 5.2.8 Computer Implementation & Examples ...................................................... 42 5.2.9 Numerical Integration – Newmark‟s Method ............................................. 47 5.2.10 Computer Implementation & Examples ................................................... 53 5.2.11 Problems ................................................................................................... 59 5.3 Multi-Degree-of-Freedom Systems .................................................................. 63 5.3.1 General Case (based on 2DOF) ................................................................... 63 5.3.2 Free-Undamped Vibration of 2DOF Systems ............................................. 66 5.3.3 Example of a 2DOF System ........................................................................ 68 5.3.4 Case Study – Aberfeldy Footbridge, Scotland ............................................ 73 Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 2 5.4 Continuous Structures ...................................................................................... 76 5.4.1 Exact Analysis for Beams ............................................................................ 76 5.4.2 Approximate Analysis – Bolton‟s Method .................................................. 86 5.4.3 Problems ...................................................................................................... 95 5.5 Practical Design Considerations ...................................................................... 97 5.5.1 Human Response to Dynamic Excitation .................................................... 97 5.5.2 Crowd/Pedestrian Dynamic Loading .......................................................... 99 5.5.3 Damping in Structures ............................................................................... 107 5.5.4 Design Rules of Thumb ............................................................................. 109 5.6 Appendix .......................................................................................................... 114 5.6.1 Past Exam Questions ................................................................................. 114 5.6.2 References .................................................................................................. 121 5.6.3 Amplitude Solution to Equation of Motion ............................................... 123 5.6.4 Solutions to Differential Equations ........................................................... 125 5.6.5 Important Formulae ................................................................................... 134 5.6.6 Glossary ..................................................................................................... 139 Rev. 1 Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 3 5.1 Introduction 5.1.1 Outline of Structural Dynamics Modern structures are increasingly slender and have reduced redundant strength due to improved analysis and design methods. Such structures are increasingly responsive to the manner in which loading is applied with respect to time and hence the dynamic behaviour of such structures must be allowed for in design; as well as the usual static considerations. In this context then, the word dynamic simply means “changes with time”; be it force, deflection or any other form of load effect. Examples of dynamics in structures are: - Soldiers breaking step as they cross a bridge to prevent harmonic excitation; - The Tacoma Narrows Bridge footage, failure caused by vortex shedding; - The London Millennium Footbridge: lateral synchronise excitation. (a) (after Craig 1981) (b) Figure 1.1 m k Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 4 The most basic dynamic system is the mass-spring system. An example is shown in Figure 1.1(a) along with the structural idealisation of it in Figure 1.1(b). This is known as a Single Degree-of-Freedom (SDOF) system as there is only one possible displacement: that of the mass in the vertical direction. SDOF systems are of great importance as they are relatively easily analysed mathematically, are easy to understand intuitively, and structures usually dealt with by Structural Engineers can be modelled approximately using an SDOF model (see Figure 1.2 for example). Figure 1.2 (after Craig 1981). Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 5 5.1.2 An Initial Numerical Example If we consider a spring-mass system as shown in Figure 1.3 with the properties m = 10 kg and k = 100 N/m and if give the mass a deflection of 20 mm and then release it (i.e. set it in motion) we would observe the system oscillating as shown in Figure 1.3. From this figure we can identify that the time between the masses recurrence at a particular location is called the period of motion or oscillation or just the period, and we denote it T; it is the time taken for a single oscillation. The number of oscillations per second is called the frequency, denoted f, and is measured in Hertz (cycles per second). Thus we can say: 1 f T = (5.1.1) We will show (Section 2.b, equation (2.19)) for a spring-mass system that: 1 2 k f m t = (5.1.2) In our system: 1 100 0.503 Hz 2 10 f t = = And from equation (5.1.1): 1 1 1.987 secs 0.503 T f = = = Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 6 We can see from Figure 1.3 that this is indeed the period observed. Figure 1.3 To reach the deflection of 20 mm just applied, we had to apply a force of 2 N, given that the spring stiffness is 100 N/m. As noted previously, the rate at which this load is applied will have an effect of the dynamics of the system. Would you expect the system to behave the same in the following cases? - If a 2 N weight was dropped onto the mass from a very small height? - If 2 N of sand was slowly added to a weightless bucket attached to the mass? Assuming a linear increase of load, to the full 2 N load, over periods of 1, 3, 5 and 10 seconds, the deflections of the system are shown in Figure 1.4. -25 -20 -15 -10 -5 0 5 10 15 20 25 0 0.5 1 1.5 2 2.5 3 3.5 4 D i s p l a c e m e n t ( m m ) Time (s) Period T m = 10 k = 100 Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 7 Figure 1.4 Remembering that the period of vibration of the system is about 2 seconds, we can see that when the load is applied faster than the period of the system, large dynamic effects occur. Stated another way, when the frequency of loading (1, 0.3, 0.2 and 0.1 Hz for our sample loading rates) is close to, or above the natural frequency of the system (0.5 Hz in our case), we can see that the dynamic effects are large. Conversely, when the frequency of loading is less than the natural frequency of the system little dynamic effects are noticed – most clearly seen via the 10 second ramp- up of the load, that is, a 0.1 Hz load. 0 5 10 15 20 25 30 35 40 0 2 4 6 8 10 12 14 16 18 20 D e f l e c t i o n ( m m ) Time (s) Dynamic Effect of Load Application Duration 1-sec 3-sec 5-sec 10-sec Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 8 5.1.3 Case Study – Aberfeldy Footbridge, Scotland Aberfeldy footbridge is a glass fibre reinforced polymer (GFRP) cable-stayed bridge over the River Tay on Aberfeldy golf course in Aberfeldy, Scotland (Figure 1.5). Its main span is 63 m and its two side spans are 25 m, also, tests have shown that the natural frequency of this bridge is 1.52 Hz, giving a period of oscillation of 0.658 seconds. Figure 1.5: Aberfeldy Footbridge Figure 1.6: Force-time curves for walking: (a) Normal pacing. (b) Fast pacing Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 9 Footbridges are generally quite light structures as the loading consists of pedestrians; this often results in dynamically lively structures. Pedestrian loading varies as a person walks; from about 0.65 to 1.3 times the weight of the person over a period of about 0.35 seconds, that is, a loading frequency of about 2.86 Hz (Figure 1.6). When we compare this to the natural frequency of Aberfeldy footbridge we can see that pedestrian loading has a higher frequency than the natural frequency of the bridge – thus, from our previous discussion we would expect significant dynamic effects to results from this. Figure 1.7 shows the response of the bridge (at the mid-span) when a pedestrian crosses the bridge: significant dynamics are apparent. Figure 1.7: Mid-span deflection (mm) as a function of distance travelled (m). Design codes generally require the natural frequency for footbridges and other pedestrian traversed structures to be greater than 5 Hz, that is, a period of 0.2 seconds. The reasons for this are apparent after our discussion: a 0.35 seconds load application (or 2.8 Hz) is slower than the natural period of vibration of 0.2 seconds (5 Hz) and hence there will not be much dynamic effect resulting; in other words the loading may be considered to be applied statically. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 10 5.1.4 Structural Damping Look again at the frog in Figure 1.1, according to the results obtained so far which are graphed in Figures 1.3 and 1.4, the frog should oscillate indefinitely. If you have ever cantilevered a ruler off the edge of a desk and flicked it you would have seen it vibrate for a time but certainly not indefinitely; buildings do not vibrate indefinitely after an earthquake; Figure 1.7 shows the vibrations dying down quite soon after the pedestrian has left the main span of Aberfeldy bridge - clearly there is another action opposing or “damping” the vibration of structures. Figure 1.8 shows the undamped response of our model along with the damped response; it can be seen that the oscillations die out quite rapidly – this depends on the level of damping. Figure 1.8 Damping occurs in structures due to energy loss mechanisms that exist in the system. Examples are friction losses at any connection to or in the system and internal energy losses of the materials due to thermo-elasticity, hysteresis and inter-granular bonds. The exact nature of damping is difficult to define; fortunately theoretical damping has been shown to match real structures quite well. -25 -20 -15 -10 -5 0 5 10 15 20 25 0 2 4 6 8 10 12 14 16 18 20 D i s p l a c e m e n t ( m m ) Time (s) Undamped Damped m = 10 k = 100 Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 11 5.2 Single Degree-of-Freedom Systems 5.2.1 Fundamental Equation of Motion (a) (b) Figure 2.1: (a) SDOF system. (b) Free-body diagram of forces Considering Figure 2.1, the forces resisting the applied loading are considered as: - a force proportional to displacement (the usual static stiffness); - a force proportional to velocity (the damping force); - a force proportional to acceleration (D‟Alambert‟s inertial force). We can write the following symbolic equation: applied stiffness damping inertia F F F F = + + (5.2.1) Noting that: stiffness damping inertia F F F ku cu mu = ¹ ¦ = ` ¦ = ) (5.2.2) that is, stiffness × displacement, damping coefficient × velocity and mass × acceleration respectively. Note also that u represents displacement from the equilibrium position and that the dots over u represent the first and second derivatives m k u(t) c Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 12 with respect to time. Thus, noting that the displacement, velocity and acceleration are all functions of time, we have the Fundamental Equation of Motion: ( ) ( ) ( ) ( ) mu t cu t ku t F t + + = (5.2.3) In the case of free vibration, there is no forcing function and so ( ) 0 F t = which gives equation (5.2.3) as: ( ) ( ) ( ) 0 mu t cu t ku t + + = (5.2.4) We note also that the system will have a state of initial conditions: ( ) 0 0 u u = (5.2.5) ( ) 0 0 u u = (5.2.6) In equation (5.2.4), dividing across by m gives: ( ) ( ) ( ) 0 c k u t u t u t m m + + = (5.2.7) We introduce the following notation: cr c c ç = (5.2.8) 2 k m e = (5.2.9) Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 13 Or equally, k m e = (5.2.10) In which - e is called the undamped circular natural frequency and its units are radians per second (rad/s); - ç is the damping ratio which is the ratio of the damping coefficient, c, to the critical value of the damping coefficient cr c . We will see what these terms physically mean. Also, we will later see (equation (5.2.18)) that: 2 2 cr c m km e = = (5.2.11) Equations (5.2.8) and (5.2.11) show us that: 2 c m çe = (5.2.12) When equations (5.2.9) and (5.2.12) are introduced into equation (5.2.7), we get the prototype SDOF equation of motion: ( ) ( ) ( ) 2 2 0 u t u t u t çe e + + = (5.2.13) In considering free vibration only, the general solution to (5.2.13) is of a form t u Ce ì = (5.2.14) Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 14 When we substitute (5.2.14) and its derivates into (5.2.13) we get: ( ) 2 2 2 0 t Ce ì ì çeì e + + = (5.2.15) For this to be valid for all values of t, t Ce ì cannot be zero. Thus we get the characteristic equation: 2 2 2 0 ì çeì e + + = (5.2.16) the solutions to this equation are the two roots: 2 2 2 1,2 2 2 4 4 2 1 eç e ç e ì eç e ç ÷ ± ÷ = = ÷ ± ÷ (5.2.17) Therefore the solution depends on the magnitude of ç relative to 1. We have: - 1 ç < : Sub-critical damping or under-damped; Oscillatory response only occurs when this is the case – as it is for almost all structures. - 1 ç = : Critical damping; No oscillatory response occurs. - 1 ç > : Super-critical damping or over-damped; No oscillatory response occurs. Therefore, when 1 ç = , the coefficient of ( ) u t in equation (5.2.13) is, by definition, the critical damping coefficient. Thus, from equation (5.2.12): Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 15 2 cr c m e = (5.2.18) From which we get equation (5.2.11). Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 16 5.2.2 Free Vibration of Undamped Structures We will examine the case when there is no damping on the SDOF system of Figure 2.1 so 0 ç = in equations (5.2.13), (5.2.16) and (5.2.17) which then become: ( ) ( ) 2 0 u t u t e + = (5.2.19) respectively, where 1 i = ÷ . From the Appendix we see that the general solution to this equation is: ( ) cos sin u t A t B t e e = + (5.2.20) where A and B are constants to be obtained from the initial conditions of the system, equations (5.2.5) and (5.2.6). Thus, at 0 t = , from equation (5.2.20): ( ) ( ) ( ) 0 0 cos 0 sin 0 u A B u e e = + = 0 A u = (5.2.21) From equation (5.2.20): ( ) sin cos u t A t B t e e e e = ÷ + (5.2.22) And so: ( ) ( ) ( ) 0 0 0 sin 0 cos 0 u A B u B u e e e e e = ÷ + = = 0 u B e = (5.2.23) Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 17 Thus equation (5.2.20), after the introduction of equations (5.2.21) and (5.2.23), becomes: ( ) 0 0 cos sin u u t u t t e e e | | = + | \ . (5.2.24) where 0 u and 0 u are the initial displacement and velocity of the system respectively. Noting that cosine and sine are functions that repeat with period 2t , we see that ( ) 1 1 2 t T t e e t + = + (Figure 2.3) and so the undamped natural period of the SDOF system is: 2 T t e = (5.2.25) The natural frequency of the system is got from (1.1), (5.2.25) and (5.2.9): 1 1 2 2 k f T m e t t = = = (5.2.26) and so we have proved (1.2). The importance of this equation is that it shows the natural frequency of structures to be proportional to k m . This knowledge can aid a designer in addressing problems with resonance in structures: by changing the stiffness or mass of the structure, problems with dynamic behaviour can be addressed. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 18 Figure 2.2: SDOF free vibration response for (a) 0 20mm u = , 0 0 u = , (b) 0 0 u = , 0 50mm/s u = , and (c) 0 20mm u = , 0 50mm/s u = . Figure 2.2 shows the free-vibration response of a spring-mass system for various initial states of the system. It can be seen from (b) and (c) that when 0 0 u = the amplitude of displacement is not that of the initial displacement; this is obviously an important characteristic to calculate. The cosine addition rule may also be used to show that equation (5.2.20) can be written in the form: ( ) ( ) cos u t C t e u = ÷ (5.2.27) where 2 2 C A B = + and tan B A u ÷ = . Using A and B as calculated earlier for the initial conditions, we then have: ( ) ( ) cos u t t µ e u = ÷ (5.2.28) -30 -20 -10 0 10 20 30 0 0.5 1 1.5 2 2.5 3 3.5 4 D i s p l a c e m e n t ( m m ) Time (s) (a) (b) (c) m = 10 k = 100 Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 19 where µ is the amplitude of displacement and u is the phase angle, both given by: 2 2 0 0 u u µ e | | = + | \ . (5.2.29) 0 0 tan u u u e = (5.2.30) The phase angle determines the amount by which ( ) u t lags behind the function cos t e . Figure 2.3 shows the general case. Figure 2.3 Undamped free-vibration response. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 20 5.2.3 Computer Implementation & Examples Using MS Excel To illustrate an application we give the spreadsheet used to generate Figure 1.3. This can be downloaded from the course website. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 21 The input parameters (shown in red) are: - m – the mass; - k – the stiffness; - delta_t – the time step used in the response plot; - u_0 – the initial displacement, 0 u ; - v_0 – the initial velocity, 0 u . The properties of the system are then found: - w, using equation (5.2.10); - f, using equation (5.2.26); - T, using equation (5.2.26); - µ , using equation (5.2.29); - u , using equation (5.2.30). A column vector of times is dragged down, adding delta_t to each previous time value, and equation (5.2.24) (“Direct Eqn”), and equation (5.2.28) (“Cosine Eqn”) is used to calculate the response, ( ) u t , at each time value. Then the column of u-values is plotted against the column of t-values to get the plot. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 22 Using Matlab Although MS Excel is very helpful since it provides direct access to the numbers in each equation, as more concepts are introduced, we will need to use loops and create regularly-used functions. Matlab is ideally suited to these tasks, and so we will begin to use it also on the simple problems as a means to its introduction. A script to directly generate Figure 1.3, and calculate the system properties is given below: % Script to plot the undamped response of a single degree of freedom system % and to calculate its properties k = 100; % N/m - stiffness m = 10; % kg - mass delta_t = 0.1; % s - time step u0 = 0.025; % m - initial displacement v0 = 0; % m/s - initial velocity w = sqrt(k/m); % rad/s - circular natural frequency f = w/(2*pi); % Hz - natural frequency T = 1/f; % s - natural period ro = sqrt(u0^2+(v0/w)^2); % m - amplitude of vibration theta = atan(v0/(u0*w)); % rad - phase angle t = 0:delta_t:4; u = ro*cos(w*t-theta); plot(t,u); xlabel('Time (s)'); ylabel('Displacement (m)'); The results of this script are the system properties are displayed in the workspace window, and the plot is generated, as shown below: Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 23 Whilst this is quite useful, this script is limited to calculating the particular system of Figure 1.3. Instead, if we create a function that we can pass particular system properties to, then we can create this plot for any system we need to. The following function does this. Note that we do not calculate f or T since they are not needed to plot the response. Also note that we have commented the code very well, so it is easier to follow and understand when we come back to it at a later date. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 24 function [t u] = sdof_undamped(m,k,u0,v0,duration,plotflag) % This function returns the displacement of an undamped SDOF system with % parameters: % m - mass, kg % k - stiffness, N/m % u0 - initial displacement, m % v0 - initial velocity, m/s % duration - length of time of required response % plotflag - 1 or 0: whether or not to plot the response % This function returns: % t - the time vector at which the response was found % u - the displacement vector of response Npts = 1000; % compute the response at 1000 points delta_t = duration/(Npts-1); w = sqrt(k/m); % rad/s - circular natural frequency ro = sqrt(u0^2+(v0/w)^2); % m - amplitude of vibration theta = atan(v0/(u0*w)); % rad - phase angle t = 0:delta_t:duration; u = ro*cos(w*t-theta); if(plotflag == 1) plot(t,u); xlabel('Time (s)'); ylabel('Displacement (m)'); end To execute this function and replicate Figure 1.3, we call the following: [t u] = sdof_undamped(10,100,0.025,0,4,1); And get the same plot as before. Now though, we can really benefit from the function. Let‟s see the effect of an initial velocity on the response, try +0.1 m/s: [t u] = sdof_undamped(10,100,0.025,0.1,4,1); Note the argument to the function in bold – this is the +0.1 m/s initial velocity. And from this call we get the following plot: Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 25 From which we can see that the maximum response is now about 40 mm, rather than the original 25. Download the function from the course website and try some other values. 0 0.5 1 1.5 2 2.5 3 3.5 4 -0.05 -0.04 -0.03 -0.02 -0.01 0 0.01 0.02 0.03 0.04 0.05 Time (s) D i s p l a c e m e n t ( m ) Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 26 5.2.4 Free Vibration of Damped Structures Figure 2.4: Response with critical or super-critical damping When taking account of damping, we noted previously that there are 3, cases but only when 1 ç < does an oscillatory response ensue. We will not examine the critical or super-critical cases. Examples are shown in Figure 2.4. To begin, when 1 ç < (5.2.17) becomes: 1,2 d i ì eç e = ÷ ± (5.2.31) where d e is the damped circular natural frequency given by: 2 1 d e e ç = ÷ (5.2.32) which has a corresponding damped period and frequency of: Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 27 2 d d T t e = (5.2.33) 2 d d f e t = (5.2.34) The general solution to equation (5.2.14), using Euler‟s formula again, becomes: ( ) ( ) cos sin t d d u t e A t B t çe e e ÷ = + (5.2.35) and again using the initial conditions we get: 0 0 0 ( ) cos sin t d d d d u u u t e u t t çe çe e e e ÷ ( | | + = + ( | \ . ¸ ¸ (5.2.36) Using the cosine addition rule again we also have: ( ) ( ) cos t d u t e t çe µ e u ÷ = ÷ (5.2.37) In which 2 2 0 0 0 d u u u çe µ e | | + = + | \ . (5.2.38) 0 0 0 tan d u u u çe u e + = (5.2.39) Equations (5.2.35) to (5.2.39) correspond to those of the undamped case looked at previously when 0 ç = . Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 28 Figure 2.5: SDOF free vibration response for: (a) 0 ç = ; (b) 0.05 ç = ; (c) 0.1 ç = ; and (d) 0.5 ç = . Figure 2.5 shows the dynamic response of the SDOF model shown. It may be clearly seen that damping has a large effect on the dynamic response of the system – even for small values of ç . We will discuss damping in structures later but damping ratios for structures are usually in the range 0.5 to 5%. Thus, the damped and undamped properties of the systems are very similar for these structures. Figure 2.6 shows the general case of an under-critically damped system. -25 -20 -15 -10 -5 0 5 10 15 20 25 0 0.5 1 1.5 2 2.5 3 3.5 4 D i s p l a c e m e n t ( m m ) Time (s) (a) (b) (c) (d) m = 10 k = 100 varies Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 29 Figure 2.6: General case of an under-critically damped system. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 30 5.2.5 Computer Implementation & Examples Using MS Excel We can just modify our previous spreadsheet to take account of the revised equations for the amplitude (equation (5.2.38)), phase angle (equation (5.2.39)) and response (equation (5.2.37)), as well as the damped properties, to get: Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 31 Using Matlab Now can just alter our previous function and take account of the revised equations for the amplitude (equation (5.2.38)), phase angle (equation (5.2.39)) and response (equation (5.2.37)) to get the following function. This function will (of course) also work for undamped systems where 0 ç = . function [t u] = sdof_damped(m,k,xi,u0,v0,duration,plotflag) % This function returns the displacement of a damped SDOF system with % parameters: % m - mass, kg % k - stiffness, N/m % xi - damping ratio % u0 - initial displacement, m % v0 - initial velocity, m/s % duration - length of time of required response % plotflag - 1 or 0: whether or not to plot the response % This function returns: % t - the time vector at which the response was found % u - the displacement vector of response Npts = 1000; % compute the response at 1000 points delta_t = duration/(Npts-1); w = sqrt(k/m); % rad/s - circular natural frequency wd = w*sqrt(1-xi^2); % rad/s - damped circular frequency ro = sqrt(u0^2+((v0+xi*w*u0)/wd)^2); % m - amplitude of vibration theta = atan((v0+u0*xi*w)/(u0*w)); % rad - phase angle t = 0:delta_t:duration; u = ro*exp(-xi*w.*t).*cos(w*t-theta); if(plotflag == 1) plot(t,u); xlabel('Time (s)'); ylabel('Displacement (m)'); end Let‟s apply this to our simple example again, for 0.1 ç = : [t u] = sdof_damped(10,100,0.1,0.025,0,4,1); To get: Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 32 To plot Figure 2.5, we just call out function several times (without plotting it each time), save the response results and then plot all together: xi = [0,0.05,0.1,0.5]; for i = 1:length(xi) [t u(i,:)] = sdof_damped(10,100,xi(i),0.025,0,4,0); end plot(t,u); xlabel('Time (s)'); ylabel('Displacement (m)'); legend('Damping: 0%','Damping: 5%','Damping: 10%','Damping: 50%'); 0 0.5 1 1.5 2 2.5 3 3.5 4 -0.02 -0.01 0 0.01 0.02 0.03 Time (s) D i s p l a c e m e n t ( m ) 0 0.5 1 1.5 2 2.5 3 3.5 4 -0.03 -0.02 -0.01 0 0.01 0.02 0.03 Time (s) D i s p l a c e m e n t ( m ) Damping: 0% Damping: 5% Damping: 10% Damping: 50% Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 33 5.2.6 Estimating Damping in Structures Examining Figure 2.6, we see that two successive peaks, n u and n m u + , m cycles apart, occur at times nT and ( ) n m T + respectively. Using equation (5.2.37) we can get the ratio of these two peaks as: 2 exp n n m d u m u tçe e + | | = | \ . (5.2.40) where ( ) exp x x e ÷ . Taking the natural log of both sides we get the logarithmic decrement of damping, o , defined as: ln 2 n n m d u m u e o tç e + = = (5.2.41) for low values of damping, normal in structural engineering, we can approximate this: 2m o tç ~ (5.2.42) thus, ( ) exp 2 1 2 n n m u e m m u o tç tç + = ~ ~ + (5.2.43) and so, Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 34 2 n n m n m u u m u ç t + + ÷ ~ (5.2.44) This equation can be used to estimate damping in structures with light damping ( 0.2 ç < ) when the amplitudes of peaks m cycles apart is known. A quick way of doing this, known as the Half-Amplitude Method, is to count the number of peaks it takes to halve the amplitude, that is 0.5 n m n u u + = . Then, using (5.2.44) we get: 0.11 m ç ~ when 0.5 n m n u u + = (5.2.45) Further, if we know the amplitudes of two successive cycles (and so 1 m= ), we can find the amplitude after p cycles from two instances of equation (5.2.43): 1 p n n p n n u u u u + + | | = | \ . (5.2.46) Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 35 5.2.7 Response of an SDOF System Subject to Harmonic Force Figure 2.7: SDOF undamped system subjected to harmonic excitation So far we have only considered free vibration; the structure has been set vibrating by an initial displacement for example. We will now consider the case when a time varying load is applied to the system. We will confine ourselves to the case of harmonic or sinusoidal loading though there are obviously infinitely many forms that a time-varying load may take – refer to the references (Appendix) for more. To begin, we note that the forcing function ( ) F t has excitation amplitude of 0 F and an excitation circular frequency of O and so from the fundamental equation of motion (5.2.3) we have: 0 ( ) ( ) ( ) sin mu t cu t ku t F t + + = O (5.2.47) The solution to equation (5.2.47) has two parts: - The complementary solution, similar to (5.2.35), which represents the transient response of the system which damps out by ( ) exp t çe ÷ . The transient response may be thought of as the vibrations caused by the initial application of the load. - The particular solution, ( ) p u t , representing the steady-state harmonic response of the system to the applied load. This is the response we will be interested in as it will account for any resonance between the forcing function and the system. m k u(t) c Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 36 The complementary solution to equation (5.2.47) is simply that of the damped free vibration case studied previously. The particular solution to equation (5.2.47) is developed in the Appendix and shown to be: ( ) ( ) sin p u t t µ u = O ÷ (5.2.48) In which ( ) ( ) 1 2 2 2 2 0 1 2 F k µ | ç| ÷ ( = ÷ + ¸ ¸ (5.2.49) 2 2 tan 1 ç| u | = ÷ (5.2.50) where the phase angle is limited to 0 u t < < and the ratio of the applied load frequency to the natural undamped frequency is: | e O = (5.2.51) the maximum response of the system will come at ( ) sin 1 t u O ÷ = and dividing (5.2.48) by the static deflection 0 F k we can get the dynamic amplification factor (DAF) of the system as: ( ) ( ) 1 2 2 2 2 DAF 1 2 D | ç| ÷ ( ÷ = ÷ + ¸ ¸ (5.2.52) At resonance, when e O= , we then have: Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 37 1 1 2 D | ç = = (5.2.53) Figure 2.8 shows the effect of the frequency ratio | on the DAF. Resonance is the phenomenon that occurs when the forcing frequency coincides with that of the natural frequency, 1 | = . It can also be seen that for low values of damping, normal in structures, very high DAFs occur; for example if 0.02 ç = then the dynamic amplification factor will be 25. For the case of no damping, the DAF goes to infinity - theoretically at least; equation (5.2.53). Figure 2.8: Variation of DAF with damping and frequency ratios. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 38 The phase angle also helps us understand what is occurring. Plotting equation (5.2.50) against | for a range of damping ratios shows: Figure 2.9: Variation of phase angle with damping and frequency ratios. Looking at this then we can see three regions: - 1 | << : the force is slowly varying and u is close to zero. This means that the response (i.e. displacement) is in phase with the force: for example, when the force acts to the right, the system displaces to the right. - 1 | >> : the force is rapidly varying and u is close to 180°. This means that the force is out of phase with the system: for example, when the force acts to the right, the system is displacing to the left. - 1 | = : the forcing frequency is equal to the natural frequency, we have resonance and 90 u = . Thus the displacement attains its peak as the force is zero. 0 0.5 1 1.5 2 2.5 3 0 45 90 135 180 Frequency Ratio P h a s e A n g l e ( d e g r e e s ) Damping: 0% Damping: 10% Damping: 20% Damping: 50% Damping: 100% Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 39 We can see these phenomena by plotting the response and forcing fun(5.2.54)ction together (though with normalized displacements for ease of interpretation), for different values of | . In this example we have used 0.2 ç = . Also, the three phase angles are 2 0.04, 0.25, 0.46 u t = respectively. Figure 2.10: Steady-state responses to illustrate phase angle. Note how the force and response are firstly “in sync” ( ~ 0 u ), then “halfway out of sync” ( 90 u = ) at resonance; and finally, “fully out of sync” ( ~180 u ) at high frequency ratio. 0 0.5 1 1.5 2 2.5 3 -2 0 2 D i s p . R a t i o 0 0.5 1 1.5 2 2.5 3 -2 0 2 D i s p . R a t i o 0 0.5 1 1.5 2 2.5 3 -2 0 2 D i s p . R a t i o Time Ratio (t/T) Dynamic Response Static Response | = 0.5; DAF = 1.29 | = 0.5; DAF = 2.5 | = 2.0; DAF = 0.32 Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 40 Maximum Steady-State Displacement The maximum steady-state displacement occurs when the DAF is a maximum. This occurs when the denominator of equation (5.2.52) is a minimum: ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 2 2 2 2 2 2 2 2 2 2 2 0 1 2 0 4 1 4 2 1 0 2 1 2 1 2 0 d D d d d | | ç| | | | | ç | ç| | | ç = ( ÷ + = ¸ ¸ ( ÷ ÷ + ( = ( ÷ + ¸ ¸ ÷ + + = The trivial solution to this equation of 0 | = corresponds to an applied forcing function that has zero frequency –the static loading effect of the forcing function. The other solution is: 2 1 2 | ç = ÷ (5.2.54) Which for low values of damping, 0.1 ç s approximately, is very close to unity. The corresponding maximum DAF is then given by substituting (5.2.54) into equation (5.2.52) to get: max 2 1 2 1 D ç ç = ÷ (5.2.55) Which reduces to equation (5.2.53) for 1 | = , as it should. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 41 Measurement of Natural Frequencies It may be seen from equation (5.2.50) that when 1 | = , 2 u t = ; this phase relationship allows the accurate measurements of the natural frequencies of structures. That is, we change the input frequency O in small increments until we can identify a peak response: the value of O at the peak response is then the natural frequency of the system. Example 2.1 gave the natural frequency based on this type of test. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 42 5.2.8 Computer Implementation & Examples Using MS Excel Again we modify our previous spreadsheet and include the extra parameters related to forced response. We‟ve also used some of the equations from the Appendix to show the transient, steady-sate and total response. Normally however, we are only interested in the steady-state response, which the total response approaches over time. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 43 Using Matlab First let‟s write a little function to return the DAF, since we will use it often: function D = DAF(beta,xi) % This function returns the DAF, D, associated with the parameters: % beta - the frequency ratio % xi - the damping ratio D = 1./sqrt((1-beta.^2).^2+(2*xi.*beta).^2); And another to return the phase angle (always in the region 0 u t < < ): function theta = phase(beta,xi) % This function returns the pahse angle, theta, associated with the % parameters: % beta - the frequency ratio % xi - the damping ratio theta = atan2((2*xi.*beta),(1-beta.^2)); % refers to complex plane With these functions, and modifying our previous damped response script, we have: function [t u] = sdof_forced(m,k,xi,u0,v0,F,Omega,duration,plotflag) % This function returns the displacement of a damped SDOF system with % parameters: % m - mass, kg % k - stiffness, N/m % xi - damping ratio % u0 - initial displacement, m % v0 - initial velocity, m/s % F - amplitude of forcing function, N % Omega - frequency of forcing function, rad/s % duration - length of time of required response % plotflag - 1 or 0: whether or not to plot the response % This function returns: % t - the time vector at which the response was found % u - the displacement vector of response Npts = 1000; % compute the response at 1000 points delta_t = duration/(Npts-1); w = sqrt(k/m); % rad/s - circular natural frequency wd = w*sqrt(1-xi^2); % rad/s - damped circular frequency beta = Omega/w; % frequency ratio D = DAF(beta,xi); % dynamic amplification factor ro = F/k*D; % m - amplitude of vibration theta = phase(beta,xi); % rad - phase angle Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 44 % Constants for the transient response Aconst = u0+ro*sin(theta); Bconst = (v0+u0*xi*w-ro*(Omega*cos(theta)-xi*w*sin(theta)))/wd; t = 0:delta_t:duration; u_transient = exp(-xi*w.*t).*(Aconst*cos(wd*t)+Bconst*sin(wd*t)); u_steady = ro*sin(Omega*t-theta); u = u_transient + u_steady; if(plotflag == 1) plot(t,u,'k'); hold on; plot(t,u_transient,'k:'); plot(t,u_steady,'k--'); hold off; xlabel('Time (s)'); ylabel('Displacement (m)'); legend('Total Response','Transient','Steady-State'); end Running this for the same problem as before with 0 10 N F = and 15 rad/s O= gives: [t u] = sdof_forced(10,100,0.1,0.025,0,20,15,6,1); As can be seen, the total response quickly approaches the steady-state response. 0 1 2 3 4 5 6 -0.06 -0.04 -0.02 0 0.02 0.04 0.06 Time (s) D i s p l a c e m e n t ( m ) Total Response Transient Steady-State Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 45 Next let‟s use our little DAF function to plot something similar to Figure 2.8, but this time showing the frequency ratio and maximum response from equation (5.2.54): % Script to plot DAF against Beta for different damping ratios xi = [0.0001,0.1,0.15,0.2,0.3,0.4,0.5,1.0]; beta = 0.01:0.01:3; for i = 1:length(xi) D(i,:) = DAF(beta,xi(i)); end % A new xi vector for the maxima line xi = 0:0.01:1.0; xi(end) = 0.99999; % very close to unity xi(1) = 0.00001; % very close to zero for i = 1:length(xi) betamax(i) = sqrt(1-2*xi(i)^2); Dmax(i) = DAF(betamax(i),xi(i)); end plot(beta,D); hold on; plot(betamax,Dmax,'k--'); xlabel('Frequency Ratio'); ylabel('Dynamic Amplification'); ylim([0 6]); % set y-axis limits since DAF at xi = 0 is enormous legend( 'Damping: 0%','Damping: 10%','Damping: 15%',... 'Damping: 20%','Damping: 30%','Damping: 40%',... 'Damping: 50%', 'Damping: 100%', 'Maxima'); This gives: 0 0.5 1 1.5 2 2.5 3 0 1 2 3 4 5 6 Frequency Ratio D y n a m i c A m p l i f i c a t i o n Damping: 0% Damping: 10% Damping: 15% Damping: 20% Damping: 30% Damping: 40% Damping: 50% Damping: 100% Maxima Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 46 Lastly then, using the phase function we wrote, we can generate Figure 2.9: % Script to plot phase against Beta for different damping ratios xi = [0.0001,0.1,0.2,0.5,1.0]; beta = 0.01:0.01:3; for i = 1:length(xi) T(i,:) = phase(beta,xi(i))*(180/pi); % in degrees end plot(beta,T); xlabel('Frequency Ratio'); ylabel('Phase Angle (degrees)'); ylim([0 180]); set(gca,'ytick',[0 45 90 135 180]); grid on; legend('Damping: 0%','Damping: 10%','Damping: 20%','Damping: 50%',... 'Damping: 100%','Location','SE'); Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 47 5.2.9 Numerical Integration – Newmark’s Method Introduction The loading that can be applied to a structure is infinitely variable and closed-form mathematical solutions can only be achieved for a small number of cases. For arbitrary excitation we must resort to computational methods, which aim to solve the basic structural dynamics equation, at the next time-step: 1 1 1 1 i i i i mu cu ku F + + + + + + = (5.2.56) There are three basic time-stepping approaches to the solution of the structural dynamics equations: 1. Interpolation of the excitation function; 2. Use of finite differences of velocity and acceleration; 3. An assumed variation of acceleration. We will examine one method from the third category only. However, it is an important method and is extensible to non-linear systems, as well as multi degree-of- freedom systems (MDOF). Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 48 Development of Newmark’s Method In 1959 Newmark proposed a general assumed variation of acceleration method: ( ) ( ) 1 1 1 i i i i u u t u t u ¸ ¸ + + = + ÷ A + A ( ¸ ¸ (5.2.57) ( ) ( )( ) ( ) 2 2 1 1 0.5 i i i i i u u t u t u t u | | + + ( ( = + A + ÷ A + A ¸ ¸ ¸ ¸ (5.2.58) The parameters | and ¸ define how the acceleration is assumed over the time step, t A . Usual values are 1 2 ¸ = and 1 1 6 4 | s s . For example: - Constant (average) acceleration is given by: 1 2 ¸ = and 1 4 | = ; - Linear variation of acceleration is given by: 1 2 ¸ = and 1 6 | = . The three equations presented thus far (equations (5.2.56), (5.2.57) and (5.2.58)) are sufficient to solve for the three unknown responses at each time step. However to avoid iteration, we introduce the incremental form of the equations: 1 i i i u u u + A ÷ ÷ (5.2.59) 1 i i i u u u + A ÷ ÷ (5.2.60) 1 i i i u u u + A ÷ ÷ (5.2.61) 1 i i i F F F + A ÷ ÷ (5.2.62) Thus, Newmark‟s equations can now be written as: ( ) ( ) i i i u t u t u ¸ A = A + A A (5.2.63) Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 49 ( ) ( ) ( ) 2 2 2 i i i i t u t u u t u | A A = A + + A A (5.2.64) Solving equation (5.2.64) for the unknown change in acceleration gives: ( ) ( ) 2 1 1 1 2 i i i i u u u u t t | | | A = A ÷ ÷ A A (5.2.65) Substituting this into equation (5.2.63) and solving for the unknown increment in velocity gives: ( ) 1 2 i i i i u u u t u t ¸ ¸ ¸ | | | | | A = A ÷ + A ÷ | A \ . (5.2.66) Next we use the incremental equation of motion, derived from equation (5.2.56): i i i i m u c u k u F A + A + A = A (5.2.67) And introduce equations (5.2.65) and (5.2.66) to get: ( ) ( ) ( ) 2 1 1 1 2 1 2 i i i i i i i i m u u u t t c u u t u k u F t | | | ¸ ¸ ¸ | | | ( A ÷ ÷ ( A A ( ¸ ¸ ( | | + A ÷ + A ÷ + A = A ( | A \ . ¸ ¸ (5.2.68) Collecting terms gives: Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 50 ( ) ( ) ( ) 2 1 1 1 1 2 2 i i i i m c k u t t F m c u m t c u t ¸ | | ¸ ¸ | | | | ( + + A ( A A ( ¸ ¸ ( ( | | = A + + + + A ÷ ( ( | A \ . ¸ ¸ ¸ ¸ (5.2.69) Let‟s introduce the following for ease of representation: ( ) ( ) 2 1 ˆ k m c k t t ¸ | | = + + A A (5.2.70) ( ) 1 1 ˆ 1 2 2 i i i i F F m c u m t c u t ¸ ¸ | | | | ( ( | | A = A + + + + A ÷ ( ( | A \ . ¸ ¸ ¸ ¸ (5.2.71) Which are an effective stiffness and effective force at time i. Thus equation (5.2.69) becomes: ˆ ˆ i i k u F A = A (5.2.72) Since ˆ k and ˆ i F A are known from the system properties (m, c, k); the algorithm properties (¸ , | , t A ); and the previous time-step ( i u , i u ), we can solve equation (5.2.72) for the displacement increment: ˆ ˆ i i F u k A A = (5.2.73) Once the displacement increment is known, we can solve for the velocity and acceleration increments from equations (5.2.66) and (5.2.65) respectively. And once Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 51 all the increments are known we can compute the properties at the current time-step by just adding to the values at the previous time-step, equations (5.2.59) to (5.2.61). Newmark‟s method is stable if the time-steps is about 0.1 t T A = of the system. The coefficients in equation (5.2.71) are constant (once t A is), so we can calculate these at the start as: ( ) 1 A m c t ¸ | | = + A (5.2.74) 1 1 2 2 B m t c ¸ | | | | = + A ÷ | \ . (5.2.75) Making equation (5.2.71) become: ˆ i i i i F F Au Bu A = A + + (5.2.76) Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 52 Newmark’s Algorithm 1. Select algorithm parameters, ¸ , | and t A ; 2. Initial calculations: a. Find the initial acceleration: ( ) 0 0 0 0 1 u F cu ku m = ÷ ÷ (5.2.77) b. Calculate the effective stiffness, ˆ k from equation (5.2.70); c. Calculate the coefficients for equation (5.2.71) from equations (5.2.74) and (5.2.75). 3. For each time step, i, calculate: ˆ i i i i F F Au Bu A = A + + (5.2.78) ˆ ˆ i i F u k A A = (5.2.79) ( ) 1 2 i i i i u u u t u t ¸ ¸ ¸ | | | | | A = A ÷ + A ÷ | A \ . (5.2.80) ( ) ( ) 2 1 1 1 2 i i i i u u u u t t | | | A = A ÷ ÷ A A (5.2.81) 1 i i i u u u ÷ = + A (5.2.82) 1 i i i u u u ÷ = + A (5.2.83) 1 i i i u u u ÷ = + A (5.2.84) Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 53 5.2.10 Computer Implementation & Examples Using MS Excel Based on our previous spreadsheet, we implement Newmark Integration. Download it from the course website, and see how the equations and algorithm are implemented. In the example shown, we‟ve applied a sinusoidal load of 10 N for 0.6 secs to the system we‟ve been using so far: Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 54 Using Matlab There are no shortcuts to this one. We must write a completely new function that implements the Newmark Integration algorithm as we‟ve described it: function [u ud udd] = newmark_sdof(m, k, xi, t, F, u0, ud0, plotflag) % This function computes the response of a linear damped SDOF system % subject to an arbitrary excitation. The input parameters are: % m - scalar, mass, kg % k - scalar, stiffness, N/m % xi - scalar, damping ratio % t - vector of length N, in equal time steps, s % F - vector of length N, force at each time step, N % u0 - scalar, initial displacement, m % v0 - scalar, initial velocity, m/s % plotflag - 1 or 0: whether or not to plot the response % The output is: % u - vector of length N, displacement response, m % ud - vector of length N, velocity response, m/s % udd - vector of length N, acceleration response, m/s2 % Set the Newmark Integration parameters % gamma = 1/2 always % beta = 1/6 linear acceleration % beta = 1/4 average acceleration gamma = 1/2; beta = 1/6; N = length(t); % the number of integration steps dt = t(2)-t(1); % the time step w = sqrt(k/m); % rad/s - circular natural frequency c = 2*xi*k/w; % the damping coefficient % Calulate the effective stiffness keff = k + (gamma/(beta*dt))*c+(1/(beta*dt^2))*m; % Calulate the coefficients A and B Acoeff = (1/(beta*dt))*m+(gamma/beta)*c; Bcoeff = (1/(2*beta))*m + dt*(gamma/(2*beta)-1)*c; % calulate the change in force at each time step dF = diff(F); % Set initial state u(1) = u0; ud(1) = ud0; udd(1) = (F(1)-c*ud0-k*u0)/m; % the initial acceleration for i = 1:(N-1) % N-1 since we already know solution at i = 1 dFeff = dF(i) + Acoeff*ud(i) + Bcoeff*udd(i); dui = dFeff/keff; dudi = (gamma/(beta*dt))*dui-(gamma/beta)*ud(i)+dt*(1- gamma/(2*beta))*udd(i); duddi = (1/(beta*dt^2))*dui-(1/(beta*dt))*ud(i)-(1/(2*beta))*udd(i); u(i+1) = u(i) + dui; ud(i+1) = ud(i) + dudi; Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 55 udd(i+1) = udd(i) + duddi; end if(plotflag == 1) subplot(4,1,1) plot(t,F,'k'); xlabel('Time (s)'); ylabel('Force (N)'); subplot(4,1,2) plot(t,u,'k'); xlabel('Time (s)'); ylabel('Displacement (m)'); subplot(4,1,3) plot(t,ud,'k'); xlabel('Time (s)'); ylabel('Velocity (m/s)'); subplot(4,1,4) plot(t,udd,'k'); xlabel('Time (s)'); ylabel('Acceleration (m/s2)'); end Bear in mind that most of this script is either comments or plotting commands – Newmark Integration is a fast and small algorithm, with a huge range of applications. In order to use this function, we must write a small script that sets the problem up and then calls the newmark_sdof function. The main difficulty is in generating the forcing function, but it is not that hard: % script that calls Newmark Integration for sample problem m = 10; k = 100; xi = 0.1; u0 = 0; ud0 = 0; t = 0:0.1:4.0; % set the time vector F = zeros(1,length(t)); % empty F vector % set sinusoidal force of 10 over 0.6 s Famp = 10; Tend = 0.6; i = 1; while t(i) < Tend F(i) = Famp*sin(pi*t(i)/Tend); i = i+1; end [u ud udd] = newmark_sdof(m, k, xi, t, F, u0, ud0, 1); This produces the following plot: Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 56 Explosions are often modelled as triangular loadings. Let‟s implement this for our system: 0 0.5 1 1.5 2 2.5 3 3.5 4 -5 0 5 10 Time (s) F o r c e ( N ) 0 0.5 1 1.5 2 2.5 3 3.5 4 -0.1 0 0.1 Time (s) D i s p l a c e m e n t ( m ) 0 0.5 1 1.5 2 2.5 3 3.5 4 -0.5 0 0.5 Time (s) V e l o c i t y ( m / s ) 0 0.5 1 1.5 2 2.5 3 3.5 4 -1 0 1 Time (s) A c c e l e r a t i o n ( m / s 2 ) Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 57 % script that finds explosion response m = 10; k = 100; xi = 0.1; u0 = 0; ud0 = 0; Fmax = 50; % N Tend = 0.2; % s t = 0:0.01:2.0; % set the time vector F = zeros(1,length(t)); % empty F vector % set reducing triangular force i = 1; while t(i) < Tend F(i) = Fmax*(1-t(i)/Tend); i = i+1; end [u ud udd] = newmark_sdof(m, k, xi, t, F, u0, ud0, 1); As can be seen from the following plot, even though the explosion only lasts for a brief period of time, the vibrations will take several periods to dampen out. Also notice that the acceleration response is the most sensitive – this is the most damaging to the building, as force is mass times acceleration: the structure thus undergoes massive forces, possibly leading to damage or failure. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 58 0 0.5 1 1.5 2 2.5 3 3.5 4 0 20 40 60 Time (s) F o r c e ( N ) 0 0.5 1 1.5 2 2.5 3 3.5 4 -0.2 0 0.2 Time (s) D i s p l a c e m e n t ( m ) 0 0.5 1 1.5 2 2.5 3 3.5 4 -0.5 0 0.5 Time (s) V e l o c i t y ( m / s ) 0 0.5 1 1.5 2 2.5 3 3.5 4 -5 0 5 Time (s) A c c e l e r a t i o n ( m / s 2 ) Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 59 5.2.11 Problems Problem 1 A harmonic oscillation test gave the natural frequency of a water tower to be 0.41 Hz. Given that the mass of the tank is 150 tonnes, what deflection will result if a 50 kN horizontal load is applied? You may neglect the mass of the tower. Ans: 50.2 mm Problem 2 A 3 m high, 8 m wide single-bay single-storey frame is rigidly jointed with a beam of mass 5,000 kg and columns of negligible mass and stiffness of EI c = 4.5×10 3 kNm 2 . Calculate the natural frequency in lateral vibration and its period. Find the force required to deflect the frame 25 mm laterally. Ans: 4.502 Hz; 0.222 sec; 100 kN Problem 3 An SDOF system (m = 20 kg, k = 350 N/m) is given an initial displacement of 10 mm and initial velocity of 100 mm/s. (a) Find the natural frequency; (b) the period of vibration; (c) the amplitude of vibration; and (d) the time at which the third maximum peak occurs. Ans: 0.666 Hz; 1.502 sec; 25.91 mm; 3.285 sec. Problem 4 For the frame of Problem 2, a jack applied a load of 100 kN and then instantaneously released. On the first return swing a deflection of 19.44 mm was noted. The period of motion was measured at 0.223 sec. Assuming that the stiffness of the columns cannot Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 60 change, find (a) the damping ratio; (b) the coefficient of damping; (c) the undamped frequency and period; and (d) the amplitude after 5 cycles. Ans: 0.04; 11,367 kg·s/m; 4.488 Hz; 0.2228 sec; 7.11 mm. Problem 5 From the response time-history of an SDOF system given: Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 61 (a) estimate the damped natural frequency; (b) use the half amplitude method to calculate the damping ratio; and (c) calculate the undamped natural frequency and period. Ans: 4.021 Hz; 0.05; 4.026 Hz; 0.248 sec. Problem 6 Workers‟ movements on a platform (8 × 6 m high, m = 200 kN) are causing large dynamic motions. An engineer investigated and found the natural period in sway to be 0.9 sec. Diagonal remedial ties (E = 200 kN/mm 2 ) are to be installed to reduce the natural period to 0.3 sec. What tie diameter is required? Ans: 28.1 mm. Problem 7 The frame of examples 2.2 and 2.4 has a reciprocating machine put on it. The mass of this machine is 4 tonnes and is in addition to the mass of the beam. The machine exerts a periodic force of 8.5 kN at a frequency of 1.75 Hz. (a) What is the steady- state amplitude of vibration if the damping ratio is 4%? (b) What would the steady- state amplitude be if the forcing frequency was in resonance with the structure? Ans: 2.92 mm; 26.56 mm. Problem 8 An air conditioning unit of mass 1,600 kg is place in the middle (point C) of an 8 m long simply supported beam (EI = 8×10 3 kNm 2 ) of negligible mass. The motor runs at 300 rpm and produces an unbalanced load of 120 kg. Assuming a damping ratio of 5%, determine the steady-state amplitude and deflection at C. What rpm will result in resonance and what is the associated deflection? Ans: 1.41 mm; 22.34 mm; 206.7 rpm; 36.66 mm. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 62 Problem 9 Determine the response of our example system, with initial velocity of 0.05 m/s, when acted upon by an impulse of 0.1 s duration and magnitude 10 N at time 1.0 s. Do this up for a duration of 4 s. Ans. below Problem 10 Determine the maximum responses of a water tower which is subjected to a sinusoidal force of amplitude 445 kN and frequency 30 rad/s over 0.3 secs. The tower has properties, mass 17.5 t, stiffness 17.5 MN/m and no damping. Ans. 120 mm, 3.8 m/s, 120.7 m/s 2 Problem 11 Determine the maximum response of a system (m = 1.75 t, k = 1.75 MN/m, ç = 10%) when subjected to an increasing triangular load which reaches 22.2 kN after 0.1 s. Ans. 14.6 mm, 0.39 m/s, 15.0 m/s 2 0 0.5 1 1.5 2 2.5 3 3.5 4 -0.015 -0.01 -0.005 0 0.005 0.01 0.015 0.02 0.025 Time (s) D i s p l a c e m e n t ( m ) Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 63 5.3 Multi-Degree-of-Freedom Systems 5.3.1 General Case (based on 2DOF) (a) (b) (c) Figure 3.1: (a) 2DOF system. (b) and (c) Free-body diagrams of forces Considering Figure 3.1, we can see that the forces that act on the masses are similar to those of the SDOF system but for the fact that the springs, dashpots, masses, forces and deflections may all differ in properties. Also, from the same figure, we can see the interaction forces between the masses will result from the relative deflection between the masses; the change in distance between them. For each mass, 0 x F = ¿ , hence: ( ) ( ) 1 1 1 1 1 1 2 1 2 2 1 2 1 mu cu k u c u u k u u F + + + ÷ + ÷ = (5.3.1) Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 64 ( ) ( ) 2 2 2 2 1 2 2 1 2 mu c u u k u u F + ÷ + ÷ = (5.3.2) In which we have dropped the time function indicators and allowed u A and u A to absorb the directions of the interaction forces. Re-arranging we get: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 1 1 2 2 2 1 1 2 2 2 1 2 2 1 2 2 2 1 2 2 2 2 u m u c c u c u k k u k F u m u c u c u k u k F + + + ÷ + + + ÷ = + ÷ + + ÷ + = (5.3.3) This can be written in matrix form: 1 1 1 2 2 1 1 2 2 1 1 2 2 2 2 2 2 2 2 2 0 0 m u c c c u k k k u F m u c c u k k u F + ÷ + ÷ ( ¦ ¹ ( ¦ ¹ ( ¦ ¹ ¦ ¹ + + = ´ ` ´ ` ´ ` ´ ` ( ( ( ÷ ÷ ¸ ¸ ¹ ) ¸ ¸ ¹ ) ¸ ¸ ¹ ) ¹ ) (5.3.4) Or another way: Mu+Cu+Ku =F (5.3.5) where: M is the mass matrix (diagonal matrix); u is the vector of the accelerations for each DOF; C is the damping matrix (symmetrical matrix); u is the vector of velocity for each DOF; K is the stiffness matrix (symmetrical matrix); u is the vector of displacements for each DOF; F is the load vector. Equation (5.3.5) is quite general and reduces to many forms of analysis: Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 65 Free vibration: Mu+Cu+Ku =0 (5.3.6) Undamped free vibration: Mu+Ku = 0 (5.3.7) Undamped forced vibration: Mu+Ku = F (5.3.8) Static analysis: Ku = F (5.3.9) We will restrict our attention to the case of undamped free-vibration – equation (5.3.7) - as the inclusion of damping requires an increase in mathematical complexity which would distract from our purpose. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 66 5.3.2 Free-Undamped Vibration of 2DOF Systems The solution to (5.3.7) follows the same methodology as for the SDOF case; so following that method (equation (2.42)), we propose a solution of the form: ( ) sin t e | + u =a (5.3.10) where a is the vector of amplitudes corresponding to each degree of freedom. From this we get: ( ) 2 2 sin t e e | e ÷ + = ÷ u = a u (5.3.11) Then, substitution of (5.3.10) and (5.3.11) into (5.3.7) yields: ( ) ( ) 2 sin sin t t e e | e | ÷ + + Ma +Ka = 0 (5.3.12) Since the sine term is constant for each term: 2 e ÷ ( ¸ ¸ K M a = 0 (5.3.13) We note that in a dynamics problem the amplitudes of each DOF will be non-zero, hence, = a 0 in general. In addition we see that the problem is a standard eigenvalues problem. Hence, by Cramer‟s rule, in order for (5.3.13) to hold the determinant of 2 e ÷ K M must then be zero: 2 0 e ÷ K M = (5.3.14) For the 2DOF system, we have: Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 67 ( ) 2 2 2 2 2 1 1 2 2 2 0 k k m k m k e e e ÷ + ÷ ÷ ÷ = ( ( ¸ ¸ ¸ ¸ K M = (5.3.15) Expansion of (5.3.15) leads to an equation in 2 e called the characteristic polynomial of the system. The solutions of 2 e to this equation are the eigenvalues of 2 e ÷ ( ¸ ¸ K M . There will be two solutions or roots of the characteristic polynomial in this case and an n-DOF system has n solutions to its characteristic polynomial. In our case, this means there are two values of 2 e ( 2 1 e and 2 2 e ) that will satisfy the relationship; thus there are two frequencies for this system (the lowest will be called the fundamental frequency). For each 2 n e substituted back into (5.3.13), we will get a certain amplitude vector n a . This means that each frequency will have its own characteristic displaced shape of the degrees of freedoms called the mode shape. However, we will not know the absolute values of the amplitudes as it is a free-vibration problem; hence we express the mode shapes as a vector of relative amplitudes, n φ , relative to, normally, the first value in n a . As we will see in the following example, the implication of the above is that MDOF systems vibrate, not just in the fundamental mode, but also in higher harmonics. From our analysis of SDOF systems it‟s apparent that should any loading coincide with any of these harmonics, large DAF‟s will result (Section 2.d). Thus, some modes may be critical design cases depending on the type of harmonic loading as will be seen later. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 68 5.3.3 Example of a 2DOF System The two-storey building shown (Figure 3.2) has very stiff floor slabs relative to the supporting columns. Calculate the natural frequencies and mode shapes. 3 2 4.5 10 kNm c EI = × Figure 3.2: Shear frame problem. Figure 3.3: 2DOF model of the shear frame. We will consider the free lateral vibrations of the two-storey shear frame idealised as in Figure 3.3. The lateral, or shear stiffness of the columns is: 1 2 3 6 3 6 12 2 2 12 4.5 10 3 4 10 N/m c EI k k k h k ( = = = ( ¸ ¸ × × × = = × The characteristic polynomial is as given in (5.3.15) so we have: Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 69 6 2 6 2 12 6 4 10 2 12 8 10 5000 4 10 3000 16 10 0 15 10 4.4 10 16 10 0 e e e e × ÷ × ÷ ÷ × = ( ( ¸ ¸ ¸ ¸ × ÷ × + × = This is a quadratic equation in 2 e and so can be solved using 6 15 10 a = × , 10 4.4 10 b = ÷ × and 12 16 10 c = × in the usual expression 2 2 4 2 b b ac a e ÷ ± ÷ = Hence we get 2 1 425.3 e = and 2 2 2508 e = . This may be written: 2 425.3 2508 n ¦ ¹ = ´ ` ¹ ) ω hence 20.6 50.1 n ¦ ¹ = ´ ` ¹ ) ω rad/s and 3.28 7.97 2 n t ¦ ¹ = = ´ ` ¹ ) ω f Hz To solve for the mode shapes, we will use the appropriate form of the equation of motion, equation (5.3.13): 2 e ÷ ( ¸ ¸ K M a = 0. First solve for the 2 e = ÷ ( ¸ ¸ E K M matrix and then solve Ea = 0 for the amplitudes n a . Then, form n φ . In general, for a 2DOF system, we have: 2 1 2 2 1 1 2 1 2 2 2 2 2 2 2 2 2 0 0 n n n n k k k m k k m k k k m k k m e e e + ÷ + ÷ ÷ ( ( ( = ÷ = ( ( ( ÷ ÷ ÷ ¸ ¸ ¸ ¸ ¸ ¸ E For 2 1 425.3 e = : 6 1 5.8735 4 10 4 2.7241 ÷ ( = × ( ÷ ¸ ¸ E Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 70 Hence 1 6 1 1 2 5.8735 4 0 10 4 2.7241 0 a a ÷ ¦ ¹ ( ¦ ¹ = = ´ ` ´ ` ( ÷ ¸ ¸ ¹ ) ¹ ) Ea Taking either equation, we calculate: 1 2 1 2 1 1 1 2 1 2 5.8735 4 0 0.681 1 4 2.7241 0 0.681 0.681 a a a a a a a a ÷ ÷ = ¬ = ¹ ¦ ¹ = ` ´ ` ÷ + = ¬ = ¹ ) ) φ Similarly for 2 2 2508 e = : 6 2 4.54 4 10 4 3.524 ÷ ÷ ( = × ( ÷ ÷ ¸ ¸ E Hence, again taking either equation, we calculate: 1 2 1 2 2 1 1 2 1 2 4.54 4 0 0.881 1 4 3.524 0 0.881 0.881 a a a a a a a a ÷ ÷ ÷ = ¬ = ÷ ¹ ¦ ¹ = ` ´ ` ÷ ÷ = ¬ = ÷ ÷ ¹ ) ) φ The complete solution may be given by the following two matrices which are used in further analysis for more complicated systems. 2 425.3 2508 n ¦ ¹ = ´ ` ¹ ) ω and 1 1 1.468 1.135 ( = ( ÷ ¸ ¸ Φ For our frame, we can sketch these two frequencies and associated mode shapes: Figure 3.4. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 71 Figure 3.4: Mode shapes and frequencies of the example frame. Larger and more complex structures will have many degrees of freedom and hence many natural frequencies and mode shapes. There are different mode shapes for different forms of deformation; torsional, lateral and vertical for example. Periodic loads acting in these directions need to be checked against the fundamental frequency for the type of deformation; higher harmonics may also be important. As an example; consider a 2DOF idealisation of a cantilever which assumes stiffness proportional to the static deflection at 0.5L and L as well as half the cantilever mass „lumped‟ at the midpoint and one quarter of it lumped at the tip. The mode shapes are shown in Figure 3.5. In Section 4(a) we will see the exact mode shape for this – it is clear that the approximation is rough; but, with more DOFs it will approach a better solution. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 72 Figure 3.5: Lumped mass, 2DOF idealisation of a cantilever. Mode 1 Mode 2 Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 73 5.3.4 Case Study – Aberfeldy Footbridge, Scotland Returning to the case study in Section 1, we will look at the results of some research conducted into the behaviour of this bridge which forms part of the current research into lateral synchronise excitation discovered on the London Millennium footbridge. This is taken from a paper by Dr. Paul Archbold, formerly of University College Dublin. Mode Mode Type Measured Frequency (Hz) Predicted Frequency (Hz) 1 L1 0.98 1.14 +16% 2 V1 1.52 1.63 +7% 3 V2 1.86 1.94 +4% 4 V3 2.49 2.62 +5% 5 L2 2.73 3.04 +11% 6 V4 3.01 3.11 +3% 7 V5 3.50 3.63 +4% 8 V6 3.91 4.00 +2% 9 T1 3.48 4.17 20% 10 V7 4.40 4.45 +1% 11 V8 4.93 4.90 -1% 12 T2 4.29 5.20 +21% 13 L3 5.72 5.72 +0% 14 T3 5.72 6.07 +19% Table 1: Modal frequencies Figure 3.6: Undeformed shape Table 1 gives the first 14 mode and associated frequencies from both direct measurements of the bridge and from finite-element modelling of it. The type of Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 74 mode is also listed; L is lateral, V is vertical and T is torsional. It can be seen that the predicted frequencies differ slightly from the measured; however, the modes have been estimated in the correct sequence and there may be some measurement error. We can see now that (from Section 1) as a person walks at about 2.8 Hz, there are a lot of modes that may be excited by this loading. Also note that the overall fundamental mode is lateral – this was the reason that this bridge has been analysed – it is similar to the Millennium footbridge in this respect. Figure 1.7 illustrates the dynamic motion due to a person walking on this bridge – this is probably caused by the third or fourth mode. Several pertinent mode shapes are given in Figure 3.7. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 75 Mode 1: 1 st Lateral mode 1.14 Hz Mode 2: 1 st Vertical mode 1.63 Hz Mode 3: 2 nd Vertical mode 1.94 Hz Mode 9: 1 st Torsional mode 4.17 Hz Figure 3.7: Various Modes of Aberfeldy footbridge. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 76 5.4 Continuous Structures 5.4.1 Exact Analysis for Beams General Equation of Motion Figure 4.1: Basic beam subjected to dynamic loading: (a) beam properties and coordinates; (b) resultant forces acting on the differential element. In examining Figure 4.1, as with any continuous structure, it may be seen that any differential element will have an associated stiffness and deflection – which changes with time – and hence a different acceleration. Thus, any continuous structure has an infinite number of degrees of freedom. Discretization into an MDOF structure is certainly an option and is the basis for finite-element dynamic analyses; the more DOF‟s used the more accurate the model (Section 3.b). For some basic structures Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 77 though, the exact behaviour can be explicitly calculated. We will limit ourselves to free-undamped vibration of beams that are thin in comparison to their length. A general expression can be derived and from this, several usual cases may be established. Figure 4.2: Instantaneous dynamic deflected position. Consider the element A of Figure 4.1(b); , hence: (5.4.1) after having cancelled the common shear term. The resultant transverse inertial force is (mass × acceleration; assuming constant mass): (5.4.2) Thus we have, after dividing by the common term: (5.4.3) 0 y F = ¿ ( ) ( ) ( ) , , , 0 I V x t p x t dx dx f x t dx x c ÷ ÷ = c ( ) , V x t ( ) ( ) 2 2 , , I v x t f x t dx mdx t c = c dx ( ) ( ) ( ) 2 2 , , , V x t v x t p x t m x t c c = ÷ c c Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 78 which, with no acceleration, is the usual static relationship between shear force and applied load. By taking moments about the point A on the element, and dropping second order and common terms, we get the usual expression: (5.4.4) Differentiating this with respect to and substituting into (5.4.3), in addition to the relationship (which assumes that the beam is of constant stiffness): (5.4.5) With free vibration this is: (5.4.6) ( ) ( ) , , M x t V x t x c = c x 2 2 v M EI x c = c ( ) ( ) ( ) 4 2 4 2 , , , v x t v x t EI m p x t x t c c + = c c ( ) ( ) 4 2 4 2 , , 0 v x t v x t EI m x t c c + = c c Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 79 General Solution for Free-Undamped Vibration Examination of equation (5.4.6) yields several aspects: - It is separated into spatial ( ) and temporal ( ) terms and we may assume that the solution is also; - It is a fourth-order differential in ; hence we will need four spatial boundary conditions to solve – these will come from the support conditions at each end; - It is a second order differential in and so we will need two temporal initial conditions to solve – initial deflection and velocity at a point for example. To begin, assume the solution is of a form of separated variables: (5.4.7) where will define the deformed shape of the beam and the amplitude of vibration. Inserting the assumed solution into (5.4.6) and collecting terms we have: (5.4.8) This follows as the terms each side of the equals are functions of and separately and so must be constant. Hence, each function type (spatial or temporal) is equal to and so we have: (5.4.9) (5.4.10) x t x t ( ) ( ) ( ) , v x t x Y t | = ( ) x | ( ) Y t ( ) ( ) ( ) ( ) 4 2 2 4 2 1 1 constant x Y t EI m x x Y t t | e | c c = ÷ = = c c x t 2 e ( ) ( ) 4 2 4 x EI m x x | e | c = c ( ) ( ) 2 0 Y t Y t e + = Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 80 Equation (5.4.10) is the same as for an SDOF system (equation (2.4)) and so the solution must be of the same form (equation (2.17)): (5.4.11) In order to evaluate we will use equation (5.4.9) and we introduce: 2 4 m EI e o = (5.4.12) And assuming a solution of the form , substitution into (5.4.9) gives: ( ) ( ) 4 4 exp 0 s G sx o ÷ = (5.4.13) There are then four roots for and when each is put into (5.4.13) and added we get: ( ) ( ) ( ) ( ) ( ) 1 2 3 4 exp exp exp exp x G i x G i x G x G x | o o o o = + ÷ + + ÷ (5.4.14) In which the ‟s may be complex constant numbers, but, by using Euler‟s expressions for cos, sin, sinh and cosh we get: ( ) ( ) ( ) ( ) ( ) 1 2 3 4 sin cos sinh cosh x A x A x A x A x | o o o o = + + + ÷ (5.4.15) where the ‟s are now real constants; three of which may be evaluated through the boundary conditions; the fourth however is arbitrary and will depend on . ( ) 0 0 cos sin Y Y t Y t t e e e | | = + | \ . e ( ) exp( ) x G sx | = s G A e Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 81 Simply-supported Beam Figure 4.3: First three mode shapes and frequency parameters for an s-s beam. The boundary conditions consist of zero deflection and bending moment at each end: ( ) ( ) 2 2 0, 0 and 0, 0 v v t EI t x c = = c (5.4.16) ( ) ( ) 2 2 , 0 and , 0 v v L t EI L t x c = = c (5.4.17) Substituting (5.4.16) into equation (5.4.14) we find . Similarly, (5.4.17) gives: ( ) ( ) 1 3 2 2 1 3 sin( ) sinh( ) 0 '' sin( ) sinh( ) 0 L A L A L L A L A L | o o | o ì o o = + = = ÷ + = (5.4.18) from which, we get two possibilities: 2 4 0 A A = = Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 82 3 1 0 2 sinh( ) 0 sin( ) A L A L o o = = (5.4.19) however, since is never zero, 3 A must be, and so the non-trivial solution must give us: sin( ) 0 L o = (5.4.20) which is the frequency equation and is only satisfied when L n ì t = . Hence, from (5.4.12) we get: 2 n n EI L m t e | | = | \ . (5.4.21) and the corresponding modes shapes are therefore: ( ) 1 sin n n x x A L t | | | = | \ . (5.4.22) where 1 A is arbitrary and normally taken to be unity. We can see that there are an infinite number of frequencies and mode shapes ( n ÷·) as we would expect from an infinite number of DOFs. The first three mode shapes and frequencies are shown in Figure 4.3. sinh( ) x ì 1 0 A = Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 83 Cantilever Beam This example is important as it describes the sway behaviour of tall buildings. The boundary conditions consist of: ( ) ( ) 0, 0 and 0, 0 v v t t x c = = c (5.4.23) ( ) ( ) 2 3 2 3 , 0 and , 0 v v EI L t EI L t x x c c = = c c (5.4.24) Which represent zero displacement and slope at the support and zero bending moment and shear at the tip. Substituting (5.4.23) into equation (5.4.14) we get 4 2 A A = ÷ and 3 1 A A = ÷ . Similarly, (5.4.24) gives: ( ) ( ) 2 2 2 2 1 2 3 4 3 3 3 3 1 2 3 4 '' sin( ) cos( ) sinh( ) cosh( ) 0 ''' cos( ) sin( ) cosh( ) sinh( ) 0 L A L A L A L A L L A L A L A L A L | o o o o o o o o | o o o o o o o o = ÷ ÷ + + = = ÷ + + + = (5.4.25) where a prime indicates a derivate of x , and so we find: ( ) ( ) ( ) ( ) 1 2 1 2 sin( ) sinh( ) cos( ) cosh( ) 0 cos( ) cosh( ) sin( ) sinh( ) 0 A L L A L L A L L A L L o o o o o o o o + + + = + + ÷ + = (5.4.26) Solving for 1 A and 2 A we find: ( ) ( )( ) ( ) ( )( ) 2 1 2 2 cos( ) cosh( ) 0 sin( ) sinh( ) sin( ) sinh( ) cos( ) cosh( ) 0 sin( ) sinh( ) sin( ) sinh( ) L L A L L L L L L A L L L L o o o o o o o o o o o o ( + = ( ÷ + ÷ + ( ¸ ¸ ( + = ( ÷ + ÷ + ( ¸ ¸ (5.4.27) Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 84 In order that neither 1 A and 2 A are zero, the expression in the brackets must be zero and we are left with the frequency equation: cos( )cosh( ) 1 0 L L o o + = (5.4.28) The mode shape is got by expressing 2 A in terms of 1 A : 2 1 sin( ) sinh( ) cos( ) cosh( ) L L A A L L o o o o + = ÷ + (5.4.29) and the modes shapes are therefore: ( ) ( ) 1 sin( ) sinh( ) sin( ) sinh( ) cosh( ) cos( ) cos( ) cosh( ) n x x x A L L x x L L o o | o o o o o o ÷ ( ( = + ( + ÷ + ( ¸ ¸ (5.4.30) where again 1 A is arbitrary and normally taken to be unity. We can see from (5.4.28) that it must be solved numerically for the corresponding values of L o The natural frequencies are then got from (5.4.21) with the substitution of L o for nt . The first three mode shapes and frequencies are shown in Figure 4.4. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 85 Figure 4.4: First three mode shapes and frequency parameters for a cantilever. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 86 5.4.2 Approximate Analysis – Bolton’s Method We will now look at a simplified method that requires an understanding of dynamic behaviour but is very easy to implement. The idea is to represent, through various manipulations of mass and stiffness, any complex structure as a single SDOF system which is easily solved via an implementation of equation (1.2): 1 2 E E K f M t = (5.4.31) in which we have equivalent SDOF stiffness and mass terms. Consider a mass-less cantilever which carries two different masses, Figure 4.5: Figure 4.5: Equivalent dynamic mass distribution for a cantilever. The end deflection of a cantilever loaded at its end by a force P is well known to be 3 3 PL EI and hence the stiffness is 3 3EI L . Therefore, the frequencies of the two cantilevers of Figure 4.5 are: 1 3 1 1 3 2 EI f M x t = (5.4.32) 3 1 3 ; 2 E E EI f M L t = (5.4.33) Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 87 And so, if the two frequencies are to be equal, and considering 1 M as the mass of a small element dx when the mass per metre is m, the corresponding part of E M is: 3 E x dM mdx L | | = | \ . (5.4.34) and integrating: 3 0 0.25 L E x M mdx L mL | | = | \ . = } (5.4.35) Therefore the cantilever with self-mass uniformly distributed along its length vibrates at the same frequency as would the mass-less cantilever loaded with a mass one quarter its actual mass. This answer is not quite correct but is within 5%; it ignores the fact that every element affects the deflection (and hence vibration) of every other element. The answer is reasonable for design though. Figure 4.6: Equivalent dynamic mass distribution for an s-s beam Similarly for a simply supported beam, we have an expression for the deflection at a point: Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 88 ( ) 2 2 3 x Px L x EIL o ÷ = (5.4.36) and so its stiffness is: ( ) 2 2 3 x EIL K x L x = ÷ (5.4.37) Considering Figure 4.6, we see that, from (5.4.31): ( ) 2 3 2 1 3 48 E EIL EI L M x L x M = ÷ (5.4.38) and as the two frequencies are to be equal: ( ) 2 2 4 0 16 8/15 L E L x M x mdx L mL ÷ = = } (5.4.39) which is about half of the self-mass as we might have guessed. Proceeding in a similar way we can find equivalent spring stiffnesses and masses for usual forms of beams as given in Table 1. Table 4.1 however, also includes a refinement of the equivalent masses based on the known dynamic deflected shape rather than the static deflected shape. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 89 Table 4.1: Bolton‟s table for equivalent mass, stiffnesses and relative amplitudes. Figure 4.7: Effective SDOFs: (a) neglecting relative amplitude; (b) including relative amplitude. In considering continuous beams, the continuity over the supports requires all the spans to vibrate at the same frequency for each of its modes. Thus we may consider Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 90 summing the equivalent masses and stiffnesses for each span and this is not a bad approximation. It is equivalent to the SDOF model of Figure 4.7(a). But, if we allowed for the relative amplitude between the different spans, we would have the model of Figure 4.7(b) which would be more accurate – especially when there is a significant difference in the member stiffnesses and masses: long heavy members will have larger amplitudes than short stiff light members due to the amount of kinetic energy stored. Thus, the stiffness and mass of each span must be weighted by its relative amplitude before summing. Consider the following examples of the beam shown in Figure 4.8; the exact multipliers are known to be 10.30, 13.32, 17.72, 21.67, 40.45, 46.10, 53.89 and 60.53 for the first eight modes. Figure 4.8: Continuous beam of Examples 1 to 3. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 91 Example 1: Ignoring relative amplitude and refined M E From Table 4.1, and the previous discussion: ( ) 3 48 3 101.9 E EI K L = × + ¿ ; and 8 1 3 15 2 E M mL | | = × + | \ . ¿ , and applying (5.4.31) we have: ( ) 4 1 10.82 2 EI f mL t = The multiplier in the exact answer is 10.30: an error of 5%. Example 2: Including relative amplitude and refined M E From Table 4.1 and the previous discussion, we have: 3 3 3 48 101.9 3 1 0.4108 185.9 E EI EI EI K L L L = × × + × = ¿ 3 0.4928 1 0.4299 0.4108 1.655 E M mL mL mL = × × + × = ¿ and applying (5.4.31) we have: ( ) 4 1 10.60 2 EI f mL t = The multiplier in the exact answer is 10.30: a reduced error of 2.9%. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 92 Example 3: Calculating the frequency of a higher mode Figure 4.9: Assumed mode shape for which the frequency will be found. The mode shape for calculation is shown in Figure 4.7. We can assume supports at the midpoints of each span as they do not displace in this mode shape. Hence we have seven simply supported half-spans and one cantilever half-span, so from Table 4.1 we have: ( ) ( ) ( ) ( ) 3 3 3 48 101.9 7 1 0.4108 0.5 0.5 3022.9 7 0.4928 0.5 1 0.4299 0.5 0.4108 1.813 E E EI EI K L L EI L M m L m L mL = × × + × = = × × + × = ¿ ¿ again, applying (5.4.31), we have: ( ) 4 1 40.8 2 EI f mL t = The multiplier in the exact answer is 40.45: and error of 0.9%. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 93 Mode Shapes and Frequencies Section 2.d described how the DAF is very large when a force is applied at the natural frequency of the structure; so for any structure we can say that when it is vibrating at its natural frequency it has very low stiffness – and in the case of no damping: zero stiffness. Higher modes will have higher stiffnesses but stiffness may also be recognised in one form as 1 M EI R = (5.4.40) where R is the radius of curvature and M is bending moment. Therefore, smaller stiffnesses have a larger R and larger stiffnesses have a smaller R. Similarly then, lower modes have a larger R and higher modes have a smaller R. This enables us to distinguish between modes by their frequencies. Noting that a member in single curvature (i.e. no point of contraflexure) has a larger R than a member in double curvature (1 point of contraflexure) which in turn has a larger R than a member in triple curvature (2 points of contraflexure), we can distinguish modes by deflected shapes. Figures 4.3 and 4.4 illustrate this clearly. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 94 Figure 4.10: Typical modes and reduced structures. An important fact may be deduced from Figure 4.10 and the preceding arguments: a continuous beam of any number of identical spans has the same fundamental frequency as that of one simply supported span: symmetrical frequencies are similarly linked. Also, for non-identical spans, symmetry may exist about a support and so reduced structures may be used to estimate the frequencies of the total structure; reductions are shown in Figure 4.10(b) and (d) for symmetrical and anti- symmetrical modes. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 95 5.4.3 Problems Problem 1 Calculate the first natural frequency of a simply supported bridge of mass 7 tonnes with a 3 tonne lorry at its quarter point. It is known that a load of 10 kN causes a 3 mm deflection. Ans.: 3.95Hz. Problem 2 Calculate the first natural frequency of a 4 m long cantilever (EI = 4,320 kNm 2 ) which carries a mass of 500 kg at its centre and has self weight of 1200 kg. Ans.: 3.76 Hz. Problem 3 What is the fundamental frequency of a 3-span continuous beam of spans 4, 8 and 5 m with constant EI and m? What is the frequency when EI = 6×10 3 kNm 2 and m = 150 kg/m? Ans.: 6.74 Hz. Problem 4 Calculate the first and second natural frequencies of a two-span continuous beam; fixed at A and on rollers at B and C. Span AB is 8 m with flexural stiffness of 2EI and a mass of 1.5m. Span BC is 6 m with flexural stiffness EI and mass m per metre. What are the frequencies when EI = 4.5×10 3 kNm 2 and m = 100 kg/m? Ans.: 9.3 Hz; ? Hz. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 96 Problem 5 Calculate the first and second natural frequencies of a 4-span continuous beam of spans 4, 5, 4 and 5 m with constant EI and m? What are the frequencies when EI = 4×10 3 kNm 2 and m = 120 kg/m? What are the new frequencies when support A is fixed? Does this make it more or less susceptible to human-induced vibration? Ans.: ? Hz; ? Hz. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 97 5.5 Practical Design Considerations 5.5.1 Human Response to Dynamic Excitation Figure 5.1: Equal sensation contours for vertical vibration The response of humans to vibrations is a complex phenomenon involving the variables of the vibrations being experienced as well as the perception of it. It has Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 98 been found that the frequency range between 2 and 30 Hz is particularly uncomfortable because of resonance with major body parts (Figure 5.2). Sensation contours for vertical vibrations are shown in Figure 5.1. This graph shows that for a given frequency, as the amplitude gets larger it becomes more uncomfortable; thus it is acceleration that is governing the comfort. This is important in the design of tall buildings which sway due to wind loading: it is the acceleration that causes discomfort. This may also be realised from car-travel: at constant velocity nothing is perceptible, but, upon rapid acceleration the motion if perceived ( F ma = ). Figure 5.2: Human body response to vibration Response graphs like Figure 5.1 have been obtained for each direction of vibration but vertical motion is more uncomfortable for standing subjects; for the transverse and longitudinal cases, the difference has the effect of moving the illustrated bands up a level. Other factors are also important: the duration of exposure; waveform (which is again linked to acceleration); type of activity; and, psychological factors. An example is that low frequency exposure can result in motion sickness. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 99 5.5.2 Crowd/Pedestrian Dynamic Loading Lightweight Floors Figure 5.3: Recommended vibration limits for light floors. Vibration limits for light floors from the 1984 Canadian Standard is shown in Figure 5.2; the peak acceleration is got from: ( ) 0 0.9 2 I a f M t = (5.5.1) where I is the impulse (the area under the force time graph) and is about 70 Ns and M is the equivalent mass of the floor which is about 40% of the distributed mass. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 100 This form of approach is to be complemented by a simple analysis of an equivalent SDOF system. Also, as seen in Section 1, by keeping the fundamental frequency above 5 Hz, human loading should not be problematic. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 101 Crowd Loading This form of loading occurs in grandstands and similar structures where a large number of people are densely packed and will be responding to the same stimulus. Coordinated jumping to the beat of music, for example, can cause a DAF of about 1.97 at about 2.5 Hz. Dancing, however, normally generates frequencies of 2 – 3 Hz. Once again, by keeping the natural frequency of the structure above about 5 Hz no undue dynamic effects should be noticed. In the transverse or longitudinal directions, allowance should also be made due to the crowd-sway that may accompany some events a value of about 0.3 kN per metre of seating parallel and 0.15 kN perpendicular to the seating is an approximate method for design. Staircases can be subject to considerable dynamic forces as running up or down such may cause peak loads of up to 4-5 times the persons bodyweight over a period of about 0.3 seconds – the method for lightweight floors can be applied to this scenario. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 102 Footbridges As may be gathered from the Case Studies of the Aberfeldy Bridge, the problem is complex, however some rough guidelines are possible. Once again controlling the fundamental frequency is important; the lessons of the London Millennium and the Tacoma Narrows bridges need to be heeded though: dynamic effects may occur in any direction or mode that can be excited by any form of loading. An approximate method for checking foot bridges is the following: max st u u K¢ = (5.5.2) where st u is the static deflection under the weight of a pedestrian at the point of maximum deflection; K is a configuration factor for the type of structure (given in Table 5.1); and ¢ is the dynamic response factor got again from Figure 5.4. The maximum acceleration is then got as 2 max max u u e = (see equations (2.30) and (3.11) for example, note: 2 2 f e t = ). This is then compared to a rather simple rule that the maximum acceleration of footbridge decks should not exceed 0.5 f ± . Alternatively, BD 37/01 states: “For superstructures for which the fundamental natural frequency of vibration exceeds 5Hz for the unloaded bridge in the vertical direction and 1.5 Hz for the loaded bridge in the horizontal direction, the vibration serviceability requirement is deemed to be satisfied.” – Appendix B.1 General. Adhering to this clause (which is based on the discussion of Section 1‟s Case Study) is clearly the easiest option. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 103 Also, note from Figure 5.4 the conservative nature of the damping assumed, which, from equation (2.35) can be seen to be so based on usual values of damping in structures. Table 5.1: Configuration factors for footbridges. Table 5.2: Values of the logarithmic decrement for different bridge types. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 104 Figure 5.4: Dynamic response factor for footbridges Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 105 Design Example A simply-supported footbridge of 18 m span has a total mass of 12.6 tonnes and flexural stiffness of 3×10 5 kNm 2 . Determine the maximum amplitude of vibration and vertical acceleration caused by a 0.7 kN pedestrian walking in frequency with the bridge: the pedestrian has a stride of 0.9 m and produces an effective pulsating force of 180 N. Assume the damping to be related to 0.05 o = . Is this a comfortable bridge for the pedestrian (Figure 5.1)? The natural frequency of the bridge is, from equations (2.19) and (4.21): 8 2 3 10 3.17 Hz 2 18 12600/18 f t × = = × The static deflection is: 3 8 700 18 0.2835 mm 48 3 10 st u × = = × × Table 5.1 gives 1 K = and Figure 5.4 gives 6.8 ¢ = and so, by (5.5.2) we have: max 0.2835 1.0 6.8 1.93 mm u = × × = and so the maximum acceleration is: ( ) 2 2 3 2 max max 2 3.17 1.93 10 0.78 m/s u u e t ÷ = = × × × = We compare this to the requirement that: Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 106 max 2 0.5 0.5 0.78 0.89 m/s u f f s s s And so we deem the bridge acceptable. From Figure 5.1, with the amplitude 1.93 mm and 3.17 Hz frequency, we can see that this pedestrian will feel decidedly uncomfortable and will probably change pace to avoid this frequency of loading. The above discussion, in conjunction with Section 2.d reveals why, historically, soldiers were told to break step when crossing a slender bridge – unfortunately for some, it is more probable that this knowledge did not come from any detailed dynamic analysis; rather, bitter experience. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 107 5.5.3 Damping in Structures The importance of damping should be obvious by this stage; a slight increase may significantly reduce the DAF at resonance, equation (2.47). It was alluded to in Section 1 that the exact nature of damping is not really understood but that it has been shown that our assumption of linear viscous damping applies to the majority of structures – a notable exception is soil-structure interaction in which alternative damping models must be assumed. Table 5.3 gives some typical damping values in practice. It is notable that the materials themselves have very low damping and thus most of the damping observed comes from the joints and so can it depend on: - The materials in contact and their surface preparation; - The normal force across the interface; - Any plastic deformation in the joint; - Rubbing or fretting of the joint when it is not tightened. Table 5.4: Recommended values of damping. When the vibrations or DAF is unacceptable it is not generally acceptable to detail joints that will have higher damping than otherwise normal – there are simply too many variables to consider. Depending on the amount of extra damping needed, one Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 108 could wait for the structure to be built and then measure the damping, retro-fitting vibration isolation devices as required. Or, if the extra damping required is significant, the design of a vibration isolation device may be integral to the structure. The devices that may be installed vary; some are: - Tuned mass dampers (TMDs): a relatively small mass is attached to the primary system and is „tuned‟ to vibrate at the same frequency but to oppose the primary system; - Sloshing dampers: A large water tank is used – the sloshing motion opposes the primary system motion due to inertial effects; - Liquid column dampers: Two columns of liquid, connected at their bases but at opposite sides of the primary system slosh, in a more controlled manner to oppose the primary system motion. These are the approaches taken in many modern buildings, particularly in Japan and other earthquake zones. The Citicorp building in New York (which is famous for other reasons also) and the John Hancock building in Boston were among the first to use TMDs. In the John Hancock building a concrete block of about 300 tonnes located on the 54 th storey sits on a thin film of oil. When the building sways the inertial effects of the block mean that it moves in the opposite direction to that of the sway and so opposes the motion (relying heavily on a lack of friction). This is quite a rudimentary system compared to modern systems which have computer controlled actuators that take input from accelerometers in the building and move the block an appropriate amount. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 109 5.5.4 Design Rules of Thumb General The structure should not have any modal frequency close to the frequency of any form of periodic loading, irrespective of magnitude. This is based upon the large DAFs that may occur (Section 2.d). For normal floors of span/depth ratio less than 25 vibration is not generally a problem. Problematic floors are lightweight with spans of over about 7 m. Human loading Most forms of human loading occur at frequencies < 5 Hz (Sections 1 and 5.a) and so any structure of natural frequency greater than this should not be subject to undue dynamic excitation. Machine Loading By avoiding any of the frequencies that the machine operates at, vibrations may be minimised. The addition of either more stiffness or mass will change the frequencies the structure responds to. If the response is still not acceptable vibration isolation devices may need to be considered (Section 5.c). Approximate Frequencies The Bolton Method of Section 4.b is probably the best for those structures outside the standard cases of Section 4.a. Careful thought on reducing the size of the problem to an SDOF system usually enables good approximate analysis. Other methods are: Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 110 Structures with concentrated mass: 1 2 g f t o = Simplified rule for most structures: 18 f o = where o is the static deflection and g is the acceleration under gravity. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 111 Rayleigh Approximation A method developed by Lord Rayleigh (which is always an upper bound), based on energy methods, for estimating the lowest natural frequency of transverse beam vibration is: 2 2 2 0 2 1 2 0 L L d y EI dx dx y dm e | | | \ . = } } (5.5.3) This method can be used to estimate the fundamental frequency of MDOF systems. Considering the frame of Figure 5.5, the fundamental frequency in each direction is given by: 2 1 2 2 i i i i i i i i i i i i Qu mu g g Qu mu e = = ¿ ¿ ¿ ¿ (5.5.4) where i u is the static deflection under the dead load of the structure i Q , acting in the direction of motion, and g is the acceleration due to gravity. Thus, the first mode is approximated in shape by the static deflection under dead load. For a building, this can be applied to each of the X and Y directions to obtain the estimates of the fundamental sway modes. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 112 Figure 5.5: Rayleigh approximation for the fundamental sway frequencies of a building. Figure 5.6: Rayleigh method for approximating bridge fundamental frequencies. Likewise for a bridge, by applying the dead load in each of the vertical and horizontal directions, the fundamental lift and drag modes can be obtained. The torsional mode can also be approximated by applying the dead load at the appropriate radius of gyration and determining the resulting rotation angle, Figure 5.6. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 113 This method is particularly useful when considering the results of a detailed analysis, such as finite-element. It provides a reasonable approximate check on the output. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 114 5.6 Appendix 5.6.1 Past Exam Questions Summer 2005 Question 5 (a) The system shown in Figure Q.5(a) is known to have a static deflection of 32.7 mm for an unknown mass. 1) Find the natural frequency of the system. (10%) 2) Given that the mass is 10 kg, find the peak displacement when this mass is given an initial velocity of 500 mm/s and an initial displacement of 25 mm. (10%) 3) What time does the first positive peak occur? (10%) 4) What value of damping coefficient is required such that the amplitude after 5 oscillations is 10% of the first peak? (10%) 5) What is the peak force in the spring? (20%) (b) A cantilever riverside boardwalk has been opened to the public as shown in Figure Q.5(b); however, it was found that the structure experiences significant human- and traffic-induced vibrations. An harmonic oscillation test found the natural frequency of the structure to be 2.25 Hz. It is proposed to retro-fit braced struts at 5m spacings so that the natural period of vibration will be 9 Hz – given E = 200 kN/mm 2 and ignoring buckling effects, what area of strut is required? (40%) Ans. (a) 2.756 Hz; 38.2 mm; 0.05 s; 99 kg.s/m;114.5 N; (b) 67.5 mm 2 . FIG. Q.5(b) A B C D PIN PROPOSED STRUT 100 kg/m 2    m k FIG. Q.5(a) Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 115 Sample Paper Semester 1 2006/7 5. (a) The single-degree-of-freedom system shown in Fig. Q5(a) is known to have a static deflection of 32.7 mm for an unknown mass. (i) Find the natural frequency of the system; (2 marks) (ii) Given that the mass is 10 kg, find the peak displacement when the mass is given an initial velocity of 500 mm/s and an initial displacement of 25 mm; (2 marks) (iii) At what time does the first positive peak occur? (2 marks) (iv) What damping ratio is required such that the amplitude after 5 oscillations is 10% of the first peak? (2 marks) (v) What is the peak force in the spring? (6 marks) (b) The beam shown in Fig. Q5(b) is loaded with an air conditioning (AC) unit at its tip. The AC unit produces an unbalanced force of 100 kg which varies sinusoidally. When the speed of the AC unit is varied, it is found that the maximum steady-state deflection is 20.91 mm. Determine: (i) The damping ratio; (4 marks) (ii) The maximum deflection when the unit‟s speed is 250 rpm; (7 marks) Take the following values: • EI = 1×10 6 kNm 2 ; • Mass of the unit is 500 kg. Ans. (a) 2.756 Hz; 38.2 mm; 0.05 s; 99 kg.s/m;114.5 N; (b) ??. k m FIG. Q5(a) A B A.C. UNIT  FIG. Q5(a) Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 116 Semester 1 2006/7 5. (a) A simply-supported reinforced concrete beam, 300 mm wide × 600 mm deep spans 8 m. Its fundamental natural frequency is measured to be 6.5 Hz. In your opinion, is the beam cracked or uncracked? Use a single degree-of-freedom (SDOF) system to represent the deflection at the centre of the beam. Assume that 8/15 of the total mass of the beam contributes to the SDOF model. Take the density of reinforced concrete to be 24 kN/m 3 and E = 30 kN/mm 2 . (10 marks) (b) The beam shown in Fig. Q5(b) is loaded with an air conditioning (AC) unit at its tip. The AC unit produces an unbalanced force of 200 kg which varies sinusoidally. When the speed of the AC unit is varied, it is found that the maximum steady-state amplitude of vibration is 34.6 mm. Determine: (i) The damping ratio; (5 marks) (ii) The maximum deflection when the unit‟s speed is 100 rpm; (10 marks) Take the following values: • EI = 40×10 3 kNm 2 ; • Mass of the unit is 2000 kg; • Ignore the mass of the beam. Ans. (a) Cracked; (b) 5.1%; 41.1 mm. k m FIG. Q5(a) A B A.C. UNIT  FIG. Q5(b) Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 117 Semester 1 2007/8 QUESTION 5 (a) For the frame shown in Fig. Q5, using a single-degree-of-freedom model, determine: (i) The natural frequency and period in free vibration; (ii) An expression for the displacement at time t if member BC is displaced 20 mm and suddenly released at time t = 1 sec. (8 marks) (b) The frame is found to have 5% damping. Using appropriate approximations, what is the percentage change in deflection, 4 cycles after the frame is released, of the damped behaviour compared to the undamped behaviour? (10 marks) (c) A machine is placed on member BC which has an unbalanced force of 500 kg which varies sinusoidally. Neglecting the mass of the machine, determine: (i) the maximum displacement when the unit‟s speed is 150 rpm; (ii) the speed of the machine at resonance; (iii) the displacement at resonance. (7 marks) Note: Take the following values: - EI = 20×10 3 kNm 2 ; - M = 20 tonnes; - Consider BC as infinitely rigid. Ans.(a) 3.93 Hz; 0.254 s; 20cos[24.72(t-1)], t>1; (b) Ratio: 28.4%, change: 71.6%; (c) 0.67 mm; 236 rpm; 4.01 mm. B A C D 3EI EI       M kg FIG. Q5 Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 118 Semester 1 2008/9 QUESTION 5 The structure shown in Fig. Q5 supports a scoreboard at a sports centre. The claxton (of total mass M) which sounds the end of playing periods includes a motor which has an unbalanced mass of 100 kg which varies sinusoidally when sounded. Using a single-degree-of-freedom model for vibrations in the vertical direction, and neglecting the mass of the truss members, determine: (i) the natural frequency and period in free vibration; (ii) the damping, given that a test showed 5 cycles after a 10 mm initial displacement was imposed, the amplitude was 5.30 mm; (iii) the maximum displacement when the unit‟s speed is 1500 rpm; (iv) the speed of the machine at resonance; (v) the displacement at resonance. (25 marks) Note: Take the following values: - For all truss members: 3 20 10 kN EA= × ; - M = 5 tonnes; - Ignore the stiffness and mass of member EF. Ans. 4.9 Hz; 0.205 s; 2%; 0.008 mm; 293.2 rpm; 5.2 mm. FIG. Q5 A E C      B D CLAXTON F Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 119 Semester 1 2009/10 QUESTION 5 (a) A 3 m high, 6 m wide single-bay single-storey frame is rigidly jointed with a beam of mass 2,000 kg and columns of negligible mass and stiffness of 3 2 2.7 10 kNm EI = × . Assuming the beam to be infinitely rigid, calculate the natural frequency in lateral vibration and its period. Find the force required to deflect the frame 20 mm laterally. (10 marks) (b) A spring-mass-damper SDOF system is subject to a harmonically varying force. At resonance, the amplitude of vibration is found to be 10 mm, and at 0.80 of the resonant frequency, the amplitude is found to be 5.07 mm. Determine the damping of the system. (15 marks) Ans. 5.51 Hz, 48 kN.; 0.1. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 120 Semester 1 2010/11 QUESTION 5 (a) For the shear frame shown in Fig. Q5(a), ignoring the mass of the columns: (i) How many modes will this structure have? (ii) Sketch the mode shapes; (iii) Indicate the order of the natural frequencies associated with each mode shape (i.e. lowest to highest). (10 marks) (b) For the frame shown in Fig. Q5(b), using a single-degree-of-freedom model, determine the natural frequency and period in free vibration given that EI = 27×10 3 kNm 2 and M = 24 tonnes. If a machine is placed on member BC which has an unbalanced force of 500 kg varying sinusoidally, neglecting the mass of the machine, determine: (i) the maximum displacement when the unit‟s speed is 360 rpm; (ii) the speed of the machine at resonance; (iii) the displacement at resonance. (15 marks) Ans. 0.34 mm, 426.6 rpm, 1.02 mm. B A E F FIG. Q5(a) C D B A C D 3EI EI    M kg FIG. Q5(b) Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 121 5.6.2 References The following books/articles were referred to in the writing of these notes; particularly Clough & Penzien (1993), Smith (1988) and Bolton (1978) - these should be referred to first for more information. There is also a lot of information and software available online; the software can especially help intuitive understanding. The class notes of Mr. R. Mahony (D.I.T.) and Dr. P. Fanning (U.C.D.) were also used. - Archbold, P., (2002), “Modal Analysis of a GRP Cable-Stayed Bridge”, Proceedings of the First Symposium of Bridge Engineering Research In Ireland, Eds. C. McNally & S. Brady, University College Dublin. - Beards, C.F., (1983), Structural Vibration Analysis: modelling, analysis and damping of vibrating structures, Ellis Horwood, Chichester, England. - Bhatt, P., (1999), Structures, Longman, Harlow, England. - Bolton, A., (1978), “Natural frequencies of structures for designers”, The Structural Engineer, Vol. 56A, No. 9, pp. 245-253; Discussion: Vol. 57A, No. 6, p.202, 1979. - Bolton, A., (1969), “The natural frequencies of continuous beams”, The Structural Engineer, Vol. 47, No. 6, pp.233-240. - Case, J., Chilver, A.H. and Ross, C.T.F., (1999), Strength of Materials and Structures, 4th edn., Arnold, London. - Chopra, A.K., (2007), Dynamics of Structures – Theory and Applications to Earthquake Engineering, 3rd edn., Pearson-Prentice Hall, New Jersey. - Clough, R.W. and Penzien, J., (1993), Dynamics of Structures, 2nd edn., McGraw- Hill, New York. - Cobb, F. (2004), Structural Engineer’s Pocket Book, Elsevier, Oxford. - Craig, R.R., (1981), Structural Dynamics – An introduction to computer methods, Wiley, New York. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 122 - Ghali, A. and Neville, A.M., (1997), Structural Analysis – A unified classical and matrix approach, 4th edn., E&FN Spon, London. - Irvine, M., (1986), Structural Dynamics for the Practising Engineer, Allen & Unwin, London. - Kreyszig, E., (1993), Advanced Engineering Mathematics, 7th edn., Wiley. - Paz, M. and Leigh, W., (2004), Structural Dynamics – Theory and Computation, 5th edn., Springer, New York. - Smith, J.W., (1988), Vibration of Structures – Applications in civil engineering design, Chapman and Hall, London. Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 123 5.6.3 Amplitude Solution to Equation of Motion The solution to the equation of motion is found to be in the form: ( ) cos sin u t A t B t e e = + (5.5.5) However, we regularly wish to express it in one of the following forms: ( ) ( ) cos u t C t e o = + (5.5.6) ( ) ( ) cos u t C t e | = ÷ (5.5.7) Where 2 2 C A B = + (5.5.8) tan A B o = (5.5.9) tan B A | = (5.5.10) To arrive at this result, re-write equation (5.5.5) as: ( ) cos sin A B u t C t t C C e e ( = + ( ¸ ¸ (5.5.11) If we consider that A, B and C represent a right-angled triangle with angles o and | , then we can draw the following: Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 124 Thus: sin cos A C o | = = (5.5.12) cos sin B C o | = = (5.5.13) Introducing these into equation (5.5.11) gives two relationships: ( ) | | sin cos cos sin u t C t t o e o e = + (5.5.14) ( ) | | cos cos sin sin u t C t t | e | e = + (5.5.15) And using the well-known trigonometric identities: ( ) sin sin cos cos sin X Y X Y X Y + = + (5.5.16) ( ) cos cos cos sin sin X Y X Y X Y ÷ = + (5.5.17) Gives the two possible representations, the last of which is the one we adopt: ( ) ( ) sin u t C t e o = + (5.5.18) ( ) ( ) cos u t C t e | = ÷ (5.5.19) Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 125 5.6.4 Solutions to Differential Equations The Homogenous Equation To find the solution of: 2 2 2 0 d y k y dx + = (5.5.20) we try x y e ì = (note that this k has nothing to do with stiffness but is the conventional mathematical notation for this problem). Thus we have: 2 2 2 ; x x dy d y e e dx dx ì ì ì ì = = Substituting this into (5.5.20) gives: 2 2 0 x x e k e ì ì ì + = And so we get the characteristic equation by dividing out x e ì : 2 2 0 k ì + = From which: 2 k ì = ± ÷ Or, Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 126 1 2 ; ik ik ì ì = + = ÷ Where 1 i = ÷ . Since these are both solutions, they are both valid and the expression for y becomes: 1 2 ikx ikx y Ae Ae ÷ = + (5.5.21) In which 1 A and 2 A are constants to be determined from the initial conditions of the problem. Introducing Euler‟s equations: cos sin cos sin ikx ikx e kx i kx e kx i kx ÷ = + = ÷ (5.5.22) into (5.5.21) gives us: ( ) ( ) 1 2 cos sin cos sin y A kx i kx A kx i kx = + + ÷ Collecting terms: ( ) ( ) 1 2 1 2 cos sin y A A kx iA iA kx = + + ÷ Since the coefficients of the trigonometric functions are constants we can just write: cos sin y A kx B kx = + (5.5.23) Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 127 The Non-homogenous Equation Starting with equation (5.2.47) (repeated here for convenience): 0 ( ) ( ) ( ) sin mu t cu t ku t F t + + = O (5.5.24) We divide by m and introduce equations (5.2.10) and (5.2.12) to get: 2 0 ( ) 2 ( ) ( ) sin F u t u t u t t m çe e + + = O (5.5.25) At this point, recall that the solution to non-homogenous differential equations is made up of two parts: - The complimentary solution ( ( ) C u t ): this is the solution to the corresponding homogenous equation, which we already have (equation (5.5.23)); - The particular solution ( ( ) P u t ): particular to the function on the right hand side of equation (5.5.24), which we must now find. The final solution is the sum of the complimentary and particular solutions: ( ) ( ) ( ) C P u t u t u t = + (5.5.26) For the particular solution we try the following: ( ) sin cos P u t C t D t = O + O (5.5.27) Then we have: ( ) cos sin P u t C t D t = O O ÷O O (5.5.28) Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 128 And ( ) 2 2 sin cos P u t C t D t = ÷O O ÷O O (5.5.29) Substituting equations (5.5.27), (5.5.28) and (5.5.29) into equation (5.5.25) gives: | | | | 2 2 2 0 sin cos 2 cos sin sin cos sin C t D t C t D t F C t D t t m çe e ÷O O ÷ O O ( ¸ ¸ + O O ÷ O O + O + O = O (5.5.30) Collecting sine and cosine terms: ( ) ( ) 2 2 2 2 0 2 sin 2 cos sin C D t F C D t t m e çe çe e ( ÷ O ÷ O O ¸ ¸ ( + O + ÷ O O = O ¸ ¸ (5.5.31) For this to be valid for all t, the sine and cosine terms on both sides of the equation must be equal. Thus: ( ) 2 2 0 2 F C D m e çe ÷O ÷ O = (5.5.32) ( ) 2 2 2 0 C D çe e O + ÷O = (5.5.33) Next, divide both sides by 2 e : Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 129 2 0 2 2 1 2 F C D m ç e e e O O | | ÷ ÷ = | \ . (5.5.34) 2 2 2 1 0 C D ç e e O O | | + ÷ = | \ . (5.5.35) Introduce the frequency ratio, equation (5.2.51), | e = O , and 2 k m e = from equation (5.2.9) to get: ( ) 2 0 1 2 F C D k | ç| ÷ ÷ = (5.5.36) ( ) 2 2 1 0 C D ç| | + ÷ = (5.5.37) From equation (5.5.37), we have: ( ) 2 1 2 C D | ç| ÷ = ÷ (5.5.38) And using this in equation (5.5.36) gives: ( ) 2 2 0 1 2 2 F D k | ç| ç| ( ÷ ÷ ÷ = ( ( ¸ ¸ (5.5.39) To get: ( ) ( ) 2 2 2 0 1 2 2 F D k | ç| ç| ( ÷ + ÷ = ( ( ¸ ¸ (5.5.40) Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 130 And rearrange to get, finally: ( ) ( ) 0 2 2 2 2 1 2 F D k ç| | ç| ÷ = ÷ + (5.5.41) Now using this with equation (5.5.38), we have: ( ) ( ) ( ) 2 0 2 2 2 1 2 2 1 2 F C k | ç| ç| | ç| ( ÷ ÷ = ÷ ( ÷ + ( ¸ ¸ (5.5.42) To get, finally: ( ) ( ) ( ) 2 0 2 2 2 1 1 2 F C k | | ç| ÷ = ÷ + (5.5.43) Again we use the cosine addition rule: 2 2 C D µ = + (5.5.44) tan D C u = ÷ (5.5.45) To express the solution as: ( ) ( ) sin P u t t µ u = O ÷ (5.5.46) So we have, from equations (5.5.44), (5.5.43) and (5.5.41): Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 131 ( ) ( ) ( ) ( ) ( ) 2 2 2 0 2 2 2 2 2 2 1 2 1 2 1 2 F k | ç| µ | ç| | ç| ( ( ÷ ÷ ( ( = + ÷ + ÷ + ( ( ¸ ¸ ¸ ¸ (5.5.47) This simplifies to: ( ) ( ) ( ) ( ) 2 2 2 0 2 2 2 2 1 2 1 2 F k | ç| µ | ç| ÷ + = ( ÷ + ¸ ¸ (5.5.48) And finally we have the amplitude of displacement: ( ) ( ) 1 2 2 2 2 0 1 2 F k µ | ç| ÷ ( = ÷ + ¸ ¸ (5.5.49) To obtain the phase angle, we use equation (5.5.45) with equations (5.5.43) and (5.5.41) again to get: ( ) ( ) ( ) ( ) ( ) 0 2 2 2 2 0 2 2 2 2 1 2 tan 1 1 2 F k F k ç| | ç| u | | ç| ÷ ÷ + = ÷ ÷ ÷ + (5.5.50) Immediately we see that several terms (and the minus signs) cancel to give: 2 2 tan 1 ç| u | = ÷ (5.5.51) Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 132 Thus we have the final particular solution of equation (5.5.46) in conjunction with equations (5.5.49) and (5.5.51). As referred to previously, the total solution is the sum of the particular and complimentary solutions, which for us now becomes: ( ) ( ) ( ) C P u t u t u t = + (5.5.52) ( ) ( ) ( ) cos sin sin t d d u t e A t B t t çe e e µ u ÷ = + + O ÷ (5.5.53) Notice here that we used equation (5.2.35) since we have redefined the amplitude and phase in terms of the forcing function. To determine the unknown constants from the initial parameters, 0 u and 0 u we differentiate equation (5.5.53) to get: ( ) ( ) ( ) ( ) cos sin cos t d d d d u t e B A t A B t t çe e çe e e çe e µ u ÷ = ÷ ÷ + +O O ÷ ( ¸ ¸ (5.5.54) Now at 0 t = , we have from equations (5.5.53) and (5.5.54): ( ) 0 0 sin u u A µ u = = ÷ (5.5.55) And: ( ) ( ) 0 0 cos d u u B A e çe µ u = = ÷ +O (5.5.56) Solving for A first from equation (5.5.55) gives: 0 sin A u µ u = + (5.5.57) Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 133 And introducing this into equation (5.5.56) gives: ( ) 0 0 sin cos d u B u e µ u çe µ u = ÷ + +O (5.5.58) Multiplying out and rearranging gives: ( ) 0 0 cos sin d u B u e çe µ u çe u = ÷ + O ÷ (5.5.59) From which we get: ( ) 0 0 cos sin d u u B çe µ u çe u e + ÷ O ÷ = (5.5.60) And now we have completely defined the time history of the problem in terms of its initial parameters. Remember that: - The complimentary solution ( ( ) C u t ): represents the transient state of the system which dampens out after a period of time, as may be realized when it is seen that it is only the complementary response that is affected by the initial state ( 0 u and 0 u ) of the system, in addition to the exponentially reducing term in equation (5.5.53); - The particular solution ( ( ) P u t ): represents the steady state of the system which persists as long as the harmonic force is applied, as again may be seen from equation (5.5.53). Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 134 5.6.5 Important Formulae SDOF Systems Fundamental equation of motion ( ) ( ) ( ) ( ) mu t cu t ku t F t + + = Equation of motion for free vibration 2 ( ) 2 ( ) ( ) 0 u t u t u t çe e + + = Relationship between frequency, circular frequency, period, stiffness and mass: Fundamental frequency for an SDOF system. 1 1 2 2 k f T m e t t = = = Coefficient of damping 2 c m çe = Circular frequency 2 k m e = Damping ratio cr c c ç = Critical value of damping 2 2 cr c m km e = = General solution for free-undamped vibration ( ) ( ) cos u t t µ e u = ÷ 2 2 0 0 ; u u µ e | | = + | \ . 0 0 tan u u u e = Damped circular frequency, period and frequency 2 1 d e e ç = ÷ 2 ; d d T t e = 2 d d f e t = General solution for free-damped vibrations ( ) ( ) cos t d u t e t çe µ e u ÷ = ÷ 2 2 0 0 0 ; d u u u çe µ e | | + = + | \ . 0 0 0 tan d u u u çe u e + = Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 135 Logarithmic decrement of damping ln 2 n n m d u m u e o tç e + = = Half-amplitude method 0.11 m ç ~ when 0.5 n m n u u + = Amplitude after p-cycles 1 p n n p n n u u u u + + | | = | \ . Equation of motion for forced response (sinusoidal) 0 ( ) ( ) ( ) sin mu t cu t ku t F t + + = O General solution for forced-damped vibration response and frequency ratio ( ) ( ) sin p u t t µ u = O ÷ ( ) ( ) 1 2 2 2 2 0 1 2 ; F k µ | ç| ÷ ( = ÷ + ¸ ¸ 2 2 tan 1 ç| u | = ÷ | e O = Dynamic amplification factor (DAF) ( ) ( ) 1 2 2 2 2 DAF 1 2 D | ç| ÷ ( ÷ = ÷ + ¸ ¸ Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 136 MDOF Systems Fundamental equation of motion Mu+Cu+Ku =F Equation of motion for undamped-free vibration Mu+Ku = 0 General solution and derivates for free-undamped vibration ( ) sin t e | + u =a ( ) 2 2 sin t e e | e ÷ + = ÷ u = a u Frequency equation 2 e ÷ ( ¸ ¸ K M a = 0 General solution for 2DOF system 1 1 1 2 2 1 2 2 2 2 2 0 0 0 0 m u k k k u m u k k u + ÷ ( ¦ ¹ ( ¦ ¹ ¦ ¹ + = ´ ` ´ ` ´ ` ( ( ÷ ¹ ) ¸ ¸ ¹ ) ¸ ¸ ¹ ) Determinant of 2DOF system from Cramer‟s rule ( ) 2 2 2 2 2 1 1 2 2 2 0 k k m k m k e e e ÷ + ÷ ÷ ÷ = ( ( ¸ ¸ ¸ ¸ K M = Composite matrix 2 e = ÷ ( ¸ ¸ E K M Amplitude equation Ea = 0 Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 137 Continuous Structures Equation of motion Assumed solution for free- undamped vibrations General solution ( ) ( ) ( ) ( ) ( ) 1 2 3 4 sin cos sinh cosh x A x A x A x A x | o o o o = + + + Boundary conditions for a simply supported beam ( ) ( ) 2 2 0, 0 and 0, 0 v v t EI t x c = = c ( ) ( ) 2 2 , 0 and , 0 v v L t EI L t x c = = c Frequencies of a simply supported beam 2 n n EI L m t e | | = | \ . Mode shape or mode n: (A 1 is normally unity) ( ) 1 sin n n x x A L t | | | = | \ . Cantilever beam boundary conditions ( ) ( ) 0, 0 and 0, 0 v v t t x c = = c ( ) ( ) 2 3 2 3 , 0 and , 0 v v EI L t EI L t x x c c = = c c Frequency equation for a cantilever cos( )cosh( ) 1 0 L L o o + = Cantilever mode shapes ( ) ( ) 1 sin( ) sinh( ) sin( ) sinh( ) cos( ) cosh( ) cosh( ) cos( ) n x x L L x A L L x x o o o o | o o o o ( ÷ + ( + ( = × ( + ( ÷ ( ¸ ¸ Bolton method general equation 1 2 E E K f M t = ( ) ( ) ( ) 4 2 4 2 , , , v x t v x t EI m p x t x t c c + = c c ( ) ( ) ( ) , v x t x Y t | = Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 138 Practical Design Peak acceleration under foot-loading ( ) 0 0.9 2 I a f M t = 70 Ns I ~ 40% mass per unit area M ~ Maximum dynamic deflection max st u u K¢ = Maximum vertical acceleration 2 max max u u e = BD37/01 requirement for vertical acceleration 0.5 f ± Structural Analysis IV Chapter 5 – Structural Dynamics Dr. C. Caprani 139 5.6.6 Glossary Structural dynamics introduces many new terms and concepts so it‟s beneficial to keep track of them in one place. Fill this out as you progress through the notes. Symbol Name Units Structural Analysis IV Chapter 5 – Structural Dynamics 5.4 Continuous Structures ...................................................................................... 76 5.4.1 Exact Analysis for Beams ............................................................................ 76 5.4.2 Approximate Analysis – Bolton‟s Method .................................................. 86 5.4.3 Problems ...................................................................................................... 95 5.5 Practical Design Considerations ...................................................................... 97 5.5.1 Human Response to Dynamic Excitation .................................................... 97 5.5.2 Crowd/Pedestrian Dynamic Loading .......................................................... 99 5.5.3 Damping in Structures ............................................................................... 107 5.5.4 Design Rules of Thumb ............................................................................. 109 5.6 Appendix .......................................................................................................... 114 5.6.1 Past Exam Questions ................................................................................. 114 5.6.2 References.................................................................................................. 121 5.6.3 Amplitude Solution to Equation of Motion ............................................... 123 5.6.4 Solutions to Differential Equations ........................................................... 125 5.6.5 Important Formulae ................................................................................... 134 5.6.6 Glossary ..................................................................................................... 139 Rev. 1 2 Dr. C. Caprani Structural Analysis IV Chapter 5 – Structural Dynamics 5.1 Introduction 5.1.1 Outline of Structural Dynamics Modern structures are increasingly slender and have reduced redundant strength due to improved analysis and design methods. Such structures are increasingly responsive to the manner in which loading is applied with respect to time and hence the dynamic behaviour of such structures must be allowed for in design; as well as the usual static considerations. In this context then, the word dynamic simply means “changes with time”; be it force, deflection or any other form of load effect. Examples of dynamics in structures are:  Soldiers breaking step as they cross a bridge to prevent harmonic excitation;  The Tacoma Narrows Bridge footage, failure caused by vortex shedding;  The London Millennium Footbridge: lateral synchronise excitation. k m (a) (after Craig 1981) Figure 1.1 (b) 3 Dr. C. Caprani and structures usually dealt with by Structural Engineers can be modelled approximately using an SDOF model (see Figure 1. Figure 1. are easy to understand intuitively. 4 Dr. This is known as a Single Degree-of-Freedom (SDOF) system as there is only one possible displacement: that of the mass in the vertical direction. Caprani . An example is shown in Figure 1.2 (after Craig 1981). SDOF systems are of great importance as they are relatively easily analysed mathematically. C.1(b).2 for example).1(a) along with the structural idealisation of it in Figure 1.Structural Analysis IV Chapter 5 – Structural Dynamics The most basic dynamic system is the mass-spring system. Structural Analysis IV Chapter 5 – Structural Dynamics 5.1.3 with the properties m = 10 kg and k = 100 N/m and if give the mass a deflection of 20 mm and then release it (i. Thus we can say: f  1 T (5. Caprani .1.3.987 secs f 0. and is measured in Hertz (cycles per second).1) We will show (Section 2.1): T 1 1   1. it is the time taken for a single oscillation. C.2 An Initial Numerical Example If we consider a spring-mass system as shown in Figure 1. equation (2. denoted f.b.2) In our system: f  1 100  0.503 Hz 2 10 And from equation (5. From this figure we can identify that the time between the masses recurrence at a particular location is called the period of motion or oscillation or just the period.e.1. set it in motion) we would observe the system oscillating as shown in Figure 1.503 5 Dr.19)) for a spring-mass system that: f  1 2 k m (5.1. The number of oscillations per second is called the frequency. and we denote it T. Would you expect the system to behave the same in the following cases?  If a 2 N weight was dropped onto the mass from a very small height?  If 2 N of sand was slowly added to a weightless bucket attached to the mass? Assuming a linear increase of load.Structural Analysis IV Chapter 5 – Structural Dynamics We can see from Figure 1. Caprani . 6 Dr.5 2 2. C.5 1 1.3 To reach the deflection of 20 mm just applied. the rate at which this load is applied will have an effect of the dynamics of the system. we had to apply a force of 2 N. over periods of 1. 25 20 15 Displacement (mm) 10 5 0 0 -5 -10 -15 -20 -25 Time (s) 0. 3.3 that this is indeed the period observed. As noted previously. given that the spring stiffness is 100 N/m.5 3 3. the deflections of the system are shown in Figure 1. to the full 2 N load.5 4 k = 100 m = 10 Period T Figure 1. 5 and 10 seconds.4. we can see that when the load is applied faster than the period of the system. a 0. C.3.1 Hz for our sample loading rates) is close to. we can see that the dynamic effects are large. 0. Conversely. when the frequency of loading is less than the natural frequency of the system little dynamic effects are noticed – most clearly seen via the 10 second rampup of the load. that is. when the frequency of loading (1. 0.5 Hz in our case). 7 Dr.1 Hz load.Structural Analysis IV Chapter 5 – Structural Dynamics Dynamic Effect of Load Application Duration 40 35 30 Deflection (mm) 25 20 15 10 5 0 0 2 4 6 8 10 Time (s) 1-sec 3-sec 5-sec 10-sec 12 14 16 18 20 Figure 1.4 Remembering that the period of vibration of the system is about 2 seconds.2 and 0. Caprani . or above the natural frequency of the system (0. large dynamic effects occur. Stated another way. Structural Analysis IV Chapter 5 – Structural Dynamics 5.1.3 Case Study – Aberfeldy Footbridge, Scotland Aberfeldy footbridge is a glass fibre reinforced polymer (GFRP) cable-stayed bridge over the River Tay on Aberfeldy golf course in Aberfeldy, Scotland (Figure 1.5). Its main span is 63 m and its two side spans are 25 m, also, tests have shown that the natural frequency of this bridge is 1.52 Hz, giving a period of oscillation of 0.658 seconds. Figure 1.5: Aberfeldy Footbridge Figure 1.6: Force-time curves for walking: (a) Normal pacing. (b) Fast pacing 8 Dr. C. Caprani Structural Analysis IV Chapter 5 – Structural Dynamics Footbridges are generally quite light structures as the loading consists of pedestrians; this often results in dynamically lively structures. Pedestrian loading varies as a person walks; from about 0.65 to 1.3 times the weight of the person over a period of about 0.35 seconds, that is, a loading frequency of about 2.86 Hz (Figure 1.6). When we compare this to the natural frequency of Aberfeldy footbridge we can see that pedestrian loading has a higher frequency than the natural frequency of the bridge – thus, from our previous discussion we would expect significant dynamic effects to results from this. Figure 1.7 shows the response of the bridge (at the mid-span) when a pedestrian crosses the bridge: significant dynamics are apparent. Figure 1.7: Mid-span deflection (mm) as a function of distance travelled (m). Design codes generally require the natural frequency for footbridges and other pedestrian traversed structures to be greater than 5 Hz, that is, a period of 0.2 seconds. The reasons for this are apparent after our discussion: a 0.35 seconds load application (or 2.8 Hz) is slower than the natural period of vibration of 0.2 seconds (5 Hz) and hence there will not be much dynamic effect resulting; in other words the loading may be considered to be applied statically. 9 Dr. C. Caprani Structural Analysis IV Chapter 5 – Structural Dynamics 5.1.4 Structural Damping Look again at the frog in Figure 1.1, according to the results obtained so far which are graphed in Figures 1.3 and 1.4, the frog should oscillate indefinitely. If you have ever cantilevered a ruler off the edge of a desk and flicked it you would have seen it vibrate for a time but certainly not indefinitely; buildings do not vibrate indefinitely after an earthquake; Figure 1.7 shows the vibrations dying down quite soon after the pedestrian has left the main span of Aberfeldy bridge - clearly there is another action opposing or “damping” the vibration of structures. Figure 1.8 shows the undamped response of our model along with the damped response; it can be seen that the oscillations die out quite rapidly – this depends on the level of damping. 25 20 15 10 Displacement (mm) k = 100 5 0 0 -5 -10 2 4 6 8 10 12 14 16 18 20 m = 10 -15 -20 Undamped -25 Time (s) Damped Figure 1.8 Damping occurs in structures due to energy loss mechanisms that exist in the system. Examples are friction losses at any connection to or in the system and internal energy losses of the materials due to thermo-elasticity, hysteresis and inter-granular bonds. The exact nature of damping is difficult to define; fortunately theoretical damping has been shown to match real structures quite well. 10 Dr. C. Caprani 1) Noting that: Fstiffness  ku   Fdamping  cu  Finertia  mu   (5.2 Single Degree-of-Freedom Systems 5. Note also that u represents displacement from the equilibrium position and that the dots over u represent the first and second derivatives 11 Dr. the forces resisting the applied loading are considered as:  a force proportional to displacement (the usual static stiffness).1 Fundamental Equation of Motion u(t) k m c (a) (b) Figure 2.2. (b) Free-body diagram of forces Considering Figure 2.1.2. damping coefficient × velocity and mass × acceleration respectively.2) that is.1: (a) SDOF system. stiffness × displacement. We can write the following symbolic equation: Fapplied  Fstiffness  Fdamping  Finertia (5.  a force proportional to velocity (the damping force). Caprani .Structural Analysis IV Chapter 5 – Structural Dynamics 5. C.  a force proportional to acceleration (D‟Alambert‟s inertial force).2. Structural Analysis IV Chapter 5 – Structural Dynamics with respect to time.2. Thus.2. dividing across by m gives: u (t )  c k u (t )  u (t )  0 m m (5.2.6) In equation (5.2.2.8) 2  k m (5.3) as: mu  t   cu  t   ku  t   0 (5.4). we have the Fundamental Equation of Motion: mu  t   cu  t   ku  t   F  t  (5.9) 12 Dr.3) In the case of free vibration. there is no forcing function and so F  t   0 which gives equation (5. Caprani .4) We note also that the system will have a state of initial conditions: u0  u  0  u0  u  0  (5.2.2. C. velocity and acceleration are all functions of time.2.2.7) We introduce the following notation:  c ccr (5.5) (5. noting that the displacement. 18)) that: ccr  2m  2 km (5.2. c.Structural Analysis IV Chapter 5 – Structural Dynamics Or equally.12) When equations (5.13) In considering free vibration only.11) Equations (5.2. the general solution to (5. We will see what these terms physically mean. we will later see (equation (5.8) and (5.13) is of a form u  Cet (5.11) show us that: 2  c m (5.14) Dr.10) In which   is called the undamped circular natural frequency and its units are radians per second (rad/s).9) and (5. C.2.2.12) are introduced into equation (5.   is the damping ratio which is the ratio of the damping coefficient. we get the prototype SDOF equation of motion: u  t   2u  t    2u  t   0 (5.2.7). Caprani 13 .2.2. to the critical value of the damping coefficient ccr .2.  k m (5.2.2. Also.2.2. the critical damping coefficient.13) we get:  2  2   2  Cet  0 (5.2.13) is.2.2. when   1.    1: Super-critical damping or over-damped. Caprani . Cet cannot be zero.14) and its derivates into (5.2.12): 14 Dr.Structural Analysis IV Chapter 5 – Structural Dynamics When we substitute (5.2.17)      2  1 Therefore the solution depends on the magnitude of  relative to 1. from equation (5. No oscillatory response occurs.16) 1. the coefficient of u (t ) in equation (5. C. by definition.2. Oscillatory response only occurs when this is the case – as it is for almost all structures.2  2  4 2 2  4 2 2 (5. Thus we get the characteristic equation:  2  2   2  0 the solutions to this equation are the two roots: (5.15) For this to be valid for all values of t.    1: Critical damping. Thus. No oscillatory response occurs. We have:    1: Sub-critical damping or under-damped. Therefore.2. 2. C. 15 Dr.18) From which we get equation (5.2.11). Caprani .Structural Analysis IV Chapter 5 – Structural Dynamics 2  ccr m (5. at t  0 .2. C.6).21) From equation (5.13).2.2.Structural Analysis IV Chapter 5 – Structural Dynamics 5.2 Free Vibration of Undamped Structures We will examine the case when there is no damping on the SDOF system of Figure 2.2.2.17) which then become: u  t    2u  t   0 (5.19) respectively.2. equations (5.2.1 so   0 in equations (5.16) and (5.2.2. where i  1 .20) where A and B are constants to be obtained from the initial conditions of the system. (5.2. From the Appendix we see that the general solution to this equation is: u  t   A cos t  B sin t (5.2.2.5) and (5. from equation (5.23) Dr.20): u  t    A sin t  B cos t (5.22) And so: u  0    A sin   0   B cos   0   u0 B  u0 B u0  (5. Thus. Caprani 16 .20): u  0  A cos   0   B sin   0   u0 A  u0 (5.2. 25) and (5.21) and (5.2.2.2.3) and so the undamped natural period of the SDOF system is: T 2  (5. Noting that cosine and sine are functions that repeat with period 2 .9): f  1  1   T 2 2 k m (5. C. becomes: u  u  t   u0 cos t   0  sin t   (5. Caprani .1).20).2. 17 Dr.2.Structural Analysis IV Chapter 5 – Structural Dynamics Thus equation (5.25) The natural frequency of the system is got from (1.2.24) where u0 and u0 are the initial displacement and velocity of the system respectively. problems with dynamic behaviour can be addressed. after the introduction of equations (5.2).23). This knowledge can aid a designer in addressing problems with resonance in structures: by changing the stiffness or mass of the structure.26) and so we have proved (1. The importance of this equation is that it shows the natural frequency of structures to be proportional to k m .2. we see that   t1  T   t1  2 (Figure 2. (5.2. 2. The cosine addition rule may also be used to show that equation (5. u0  50mm/s .20) can be written in the form: u(t )  C cos t    (5. (b) u0  0 .5 4 m = 10 -20 (a) -30 Time (s) (b) (c) Figure 2. Caprani .2: SDOF free vibration response for (a) u0  20mm .2.5 2 2. and (c) u0  20mm . Using A and B as calculated earlier for the A initial conditions.27) where C  A2  B 2 and tan    B .2. Figure 2. u0  0 . It can be seen from (b) and (c) that when u0  0 the amplitude of displacement is not that of the initial displacement.5 3 3. this is obviously an important characteristic to calculate. we then have: u(t )   cos t    (5.Structural Analysis IV Chapter 5 – Structural Dynamics 30 20 k = 100 Displacement (mm) 10 0 0 -10 0. u0  50mm/s .28) 18 Dr. C.2 shows the free-vibration response of a spring-mass system for various initial states of the system.5 1 1. Caprani .2. both given by: u    u  0    2 0 2 (5.29) (5.30) tan   u0 u0 The phase angle determines the amount by which u (t ) lags behind the function cos t .3 shows the general case.Structural Analysis IV Chapter 5 – Structural Dynamics where  is the amplitude of displacement and  is the phase angle. 19 Dr. Figure 2.2. C.3 Undamped free-vibration response. Figure 2. 20 Dr.3 Computer Implementation & Examples Using MS Excel To illustrate an application we give the spreadsheet used to generate Figure 1.3.Structural Analysis IV Chapter 5 – Structural Dynamics 5. This can be downloaded from the course website.2. Caprani . C.  T.  delta_t – the time step used in the response plot.Structural Analysis IV Chapter 5 – Structural Dynamics The input parameters (shown in red) are:  m – the mass.29). A column vector of times is dragged down.  u_0 – the initial displacement. u0 . using equation (5. using equation (5. adding delta_t to each previous time value. using equation (5.2. and equation (5.2.30). C.  k – the stiffness.10).2. u0 . using equation (5.26).2. u  t  .28) (“Cosine Eqn”) is used to calculate the response. The properties of the system are then found:  w. and equation (5.2.2.   .2. Then the column of u-values is plotted against the column of t-values to get the plot. Caprani . using equation (5.24) (“Direct Eqn”).   .26).  f.  v_0 – the initial velocity. 21 Dr. at each time value. The results of this script are the system properties are displayed in the workspace window. A script to directly generate Figure 1. m = 10. u0 = 0.025.3. ro = sqrt(u0^2+(v0/w)^2). T = 1/f. and calculate the system properties is given below: % Script to plot the undamped response of a single degree of freedom system % and to calculate its properties k = 100.Structural Analysis IV Using Matlab Chapter 5 – Structural Dynamics Although MS Excel is very helpful since it provides direct access to the numbers in each equation. Caprani . Matlab is ideally suited to these tasks. and the plot is generated. delta_t = 0. % % % % % N/m kg s m m/s stiffness mass time step initial displacement initial velocity % % % % % rad/s Hz s m rad circular natural frequency natural frequency natural period amplitude of vibration phase angle w = sqrt(k/m). and so we will begin to use it also on the simple problems as a means to its introduction. ylabel('Displacement (m)').u). as more concepts are introduced. C. t = 0:delta_t:4. f = w/(2*pi). we will need to use loops and create regularly-used functions. plot(t. v0 = 0. xlabel('Time (s)'). as shown below: 22 Dr. theta = atan(v0/(u0*w)). u = ro*cos(w*t-theta).1. this script is limited to calculating the particular system of Figure 1.3. Also note that we have commented the code very well. if we create a function that we can pass particular system properties to. Instead.Structural Analysis IV Chapter 5 – Structural Dynamics Whilst this is quite useful. C. then we can create this plot for any system we need to. The following function does this. Caprani . 23 Dr. Note that we do not calculate f or T since they are not needed to plot the response. so it is easier to follow and understand when we come back to it at a later date. And get the same plot as before.amplitude of vibration % rad .1 or 0: whether or not to plot the response % This function returns: % t . C. N/m % u0 .1 m/s: [t u] = sdof_undamped(10.025.u0. ylabel('Displacement (m)').100.u).initial velocity.mass. w = sqrt(k/m).025.1. try +0. t = 0:delta_t:duration. kg % k .0.1 m/s initial velocity.length of time of required response % plotflag .stiffness.0.plotflag) % This function returns the displacement of an undamped SDOF system with % parameters: % m .Structural Analysis IV Chapter 5 – Structural Dynamics function [t u] = sdof_undamped(m. Caprani .the displacement vector of response Npts = 1000.0.4.circular natural frequency % m .initial displacement. u = ro*cos(w*t-theta). m % v0 . end % rad/s .1).3.4. Let‟s see the effect of an initial velocity on the response. theta = atan(v0/(u0*w)). if(plotflag == 1) plot(t.phase angle To execute this function and replicate Figure 1. ro = sqrt(u0^2+(v0/w)^2).k. we call the following: [t u] = sdof_undamped(10. xlabel('Time (s)').v0. m/s % duration . % compute the response at 1000 points delta_t = duration/(Npts-1).duration.the time vector at which the response was found % u . Note the argument to the function in bold – this is the +0.0.100. Now though.1). we can really benefit from the function. And from this call we get the following plot: 24 Dr. Download the function from the course website and try some other values.02 Displacement (m) 0.5 3 3. rather than the original 25.05 0.01 0 -0.01 -0.03 0.5 1 1.5 2 Time (s) 2.03 -0. Caprani .02 -0.04 -0. C.Structural Analysis IV Chapter 5 – Structural Dynamics 0.04 0. 25 Dr.05 0 0.5 4 From which we can see that the maximum response is now about 40 mm. 31) d   1   2 which has a corresponding damped period and frequency of: (5.2. To begin. when   1 (5. Caprani .2. cases but only when   1 does an oscillatory response ensue.4.17) becomes: 1.32) 26 Dr.4 Free Vibration of Damped Structures Figure 2.Structural Analysis IV Chapter 5 – Structural Dynamics 5. We will not examine the critical or super-critical cases. C. Examples are shown in Figure 2. we noted previously that there are 3.2    id where d is the damped circular natural frequency given by: (5.4: Response with critical or super-critical damping When taking account of damping.2.2. 2. C.36) Using the cosine addition rule again we also have: u(t )   et cos d t    (5.2. becomes: u(t )  et  A cos d t  B sin d t  (5.14). 27 Dr.2.38) tan   u0  u0 u0d (5.2.39) correspond to those of the undamped case looked at previously when   0 .2.37) In which  u  u0    u  0  d   2 0 2 (5.35) and again using the initial conditions we get:    u  d u0  u (t )  et u0 cos d t   0 sin d t   d     (5. Caprani .2.39) Equations (5.2.Structural Analysis IV Chapter 5 – Structural Dynamics Td  2 d (5.34) The general solution to equation (5. using Euler‟s formula again.2.2.2.33) fd  d 2 (5.35) to (5. 1.5 3 3. Figure 2.5 shows the dynamic response of the SDOF model shown. (b)   0.5 to 5%.5 2 2. Thus. It may be clearly seen that damping has a large effect on the dynamic response of the system – even for small values of  .5 1 1. C. Caprani . and (d)   0.6 shows the general case of an under-critically damped system.05 .5: SDOF free vibration response for: (a)   0 . (c)   0. We will discuss damping in structures later but damping ratios for structures are usually in the range 0.Structural Analysis IV Chapter 5 – Structural Dynamics 25 20 15 Displacement (mm) 10 5 0 0 -5 -10 -15 -20 -25 Time (s) (a) (b) (c) (d) 0. 28 Dr. Figure 2. the damped and undamped properties of the systems are very similar for these structures.5 .5 4 m = 10 k = 100 varies Figure 2. Structural Analysis IV Chapter 5 – Structural Dynamics Figure 2. C.6: General case of an under-critically damped system. Caprani . 29 Dr. 5 Computer Implementation & Examples Using MS Excel We can just modify our previous spreadsheet to take account of the revised equations for the amplitude (equation (5.Structural Analysis IV Chapter 5 – Structural Dynamics 5. C. phase angle (equation (5.2. to get: 30 Dr.2. Caprani .2.2.38)).39)) and response (equation (5.37)). as well as the damped properties. u0. function [t u] = sdof_damped(m.length of time of required response % plotflag . if(plotflag == 1) plot(t. m % v0 .1 or 0: whether or not to plot the response % This function returns: % t .Structural Analysis IV Chapter 5 – Structural Dynamics Using Matlab Now can just alter our previous function and take account of the revised equations for the amplitude (equation (5.mass.1: [t u] = sdof_damped(10. u = ro*exp(-xi*w.damped circular frequency ro = sqrt(u0^2+((v0+xi*w*u0)/wd)^2).39)) and response (equation (5.v0.1).0.37)) to get the following function. end Let‟s apply this to our simple example again. To get: 31 Dr.xi.*cos(w*t-theta).1.initial velocity. w = sqrt(k/m).duration.4. % rad/s . N/m % xi . % m .2. xlabel('Time (s)').amplitude of vibration theta = atan((v0+u0*xi*w)/(u0*w)). for   0. This function will (of course) also work for undamped systems where   0 .phase angle t = 0:delta_t:duration. m/s % duration . Caprani . kg % k .2. % rad/s .plotflag) % This function returns the displacement of a damped SDOF system with % parameters: % m . phase angle (equation (5.0. % rad .k.0.the displacement vector of response Npts = 1000.the time vector at which the response was found % u .stiffness.damping ratio % u0 .initial displacement.u). % compute the response at 1000 points delta_t = duration/(Npts-1).38)).100.circular natural frequency wd = w*sqrt(1-xi^2).2. C.*t).025. ylabel('Displacement (m)'). 5 4 To plot Figure 2.5 1 1.'Damping: 50%').0.01 0 -0.'Damping: 10%'. we just call out function several times (without plotting it each time).0.Structural Analysis IV Chapter 5 – Structural Dynamics 0.0.05.03 0.4.5 2 Time (s) 2.0.xi(i).1. ylabel('Displacement (m)').5.0.:)] = sdof_damped(10.5 1 1.02 Displacement (m) 0. save the response results and then plot all together: xi = [0.100.'Damping: 5%'.0).02 -0. 0. for i = 1:length(xi) [t u(i.01 0 -0.01 -0.5 3 3. legend('Damping: 0%'.03 Damping: Damping: Damping: Damping: 0% 5% 10% 50% 0 0.5 4 32 Dr. end plot(t.5 2 Time (s) 2.u).5 3 3. Caprani . xlabel('Time (s)').02 0 0.03 0.01 -0.02 Displacement (m) 0.5].025. C. 41) for low values of damping. we can approximate this:   2m (5.42) thus.37) we can get the ratio of these two peaks as:  2m  un  exp   un  m  d  (5.2.Structural Analysis IV Chapter 5 – Structural Dynamics 5. occur at times nT and  n  m T respectively. C. Caprani . defined as:   ln un   2m un  m d (5.6 Estimating Damping in Structures Examining Figure 2.6.2.  . normal in structural engineering. un and un m .2.2.43) and so. Using equation (5. Taking the natural log of both sides we get the logarithmic decrement of damping. un  e  exp  2m   1  2m un  m (5. 33 Dr. we see that two successive peaks.2. m cycles apart.40) where exp  x   e x .2. using (5.2. Then.2.2 ) when the amplitudes of peaks m cycles apart is known. is to count the number of peaks it takes to halve the amplitude. A quick way of doing this. C.46) 34 Dr.5un . known as the Half-Amplitude Method. Caprani . that is unm  0.45) Further.11 when unm  0. we can find the amplitude after p cycles from two instances of equation (5.2. if we know the amplitudes of two successive cycles (and so m  1 ).2.44) This equation can be used to estimate damping in structures with light damping (   0.Structural Analysis IV Chapter 5 – Structural Dynamics  un  un  m 2m un m (5.5un m (5.2.43): un  p u    n1  un  un  p (5.44) we get:  0. 2.2.7 Response of an SDOF System Subject to Harmonic Force k c u(t) m Figure 2.7: SDOF undamped system subjected to harmonic excitation So far we have only considered free vibration.35).2. This is the response we will be interested in as it will account for any resonance between the forcing function and the system. We will now consider the case when a time varying load is applied to the system. To begin. similar to (5. 35 Dr.2.47) has two parts:  The complementary solution. which represents the transient response of the system which damps out by exp  t  .3) we have: mu(t )  cu(t )  ku(t )  F0 sin t (5.2. the structure has been set vibrating by an initial displacement for example. The transient response may be thought of as the vibrations caused by the initial application of the load. we note that the forcing function F  t  has excitation amplitude of F0 and an excitation circular frequency of  and so from the fundamental equation of motion (5. representing the steady-state harmonic response of the system to the applied load. We will confine ourselves to the case of harmonic or sinusoidal loading though there are obviously infinitely many forms that a time-varying load may take – refer to the references (Appendix) for more.  The particular solution. Caprani .47) The solution to equation (5.Structural Analysis IV Chapter 5 – Structural Dynamics 5. C. u p  t  . 48) In which  1 2 2 F0  2 1   2    2     k  (5.2. Caprani .Structural Analysis IV Chapter 5 – Structural Dynamics The complementary solution to equation (5.50) tan   2 1  2 where the phase angle is limited to 0     and the ratio of the applied load frequency to the natural undamped frequency is:    (5.49) (5. C. we then have: 36 Dr.2.47) is simply that of the damped free vibration case studied previously.51) the maximum response of the system will come at sin  t     1 and dividing (5.2. The particular solution to equation (5.2.2.2.48) by the static deflection F0 k we can get the dynamic amplification factor (DAF) of the system as: 2 2 DAF  D  1   2    2     1 2 (5.52) At resonance.2. when    .47) is developed in the Appendix and shown to be: u p  t    sin  t    (5.2. 53).Structural Analysis IV Chapter 5 – Structural Dynamics D 1  1 2 (5.8 shows the effect of the frequency ratio  on the DAF. 37 Dr. the DAF goes to infinity .theoretically at least. It can also be seen that for low values of damping. normal in structures.   1. very high DAFs occur.02 then the dynamic amplification factor will be 25. Resonance is the phenomenon that occurs when the forcing frequency coincides with that of the natural frequency. equation (5.2.53) Figure 2. for example if   0. Figure 2.2. Caprani . For the case of no damping. C.8: Variation of DAF with damping and frequency ratios. 9: Variation of phase angle with damping and frequency ratios. Caprani . This means that the force is out of phase with the system: for example. when the force acts to the right.2.50) against  for a range of damping ratios shows: 180 135 Phase Angle (degrees) 90 45 Damping: Damping: Damping: Damping: Damping: 0 0. Looking at this then we can see three regions:    1 : the force is slowly varying and  is close to zero. Plotting equation (5.    1: the forcing frequency is equal to the natural frequency.    1: the force is rapidly varying and  is close to 180°. C.5 0% 10% 20% 50% 100% 3 0 Figure 2.5 1 1. Thus the displacement attains its peak as the force is zero. the system is displacing to the left. 38 Dr.5 Frequency Ratio 2 2.e. This means that the response (i. we have resonance and   90 . the system displaces to the right. when the force acts to the right.Structural Analysis IV Chapter 5 – Structural Dynamics The phase angle also helps us understand what is occurring. displacement) is in phase with the force: for example. 0.46 respectively.25.32 0 -2 0 0. for different values of  . C. Ratio  = 0.5.5 1 1. 0.Structural Analysis IV Chapter 5 – Structural Dynamics We can see these phenomena by plotting the response and forcing fun(5. In this example we have used   0. DAF = 0.5 0 -2 0 0.5 3 Figure 2.54)ction together (though with normalized displacements for ease of interpretation).5 3 2 Disp. DAF = 2. “fully out of sync” ( ~ 180 ) at high frequency ratio. Caprani .5 1 1.5 1 1. and finally.2. Ratio  = 2.5 3 2 Disp.5 Time Ratio (t/T) 2 2. then “halfway out of sync” (   90 ) at resonance. Also. DAF = 1.5 2 2.  = 0. Ratio Dynamic Response Static Response 0 -2 0 0.2 . 0.29 2 Disp.04.5. Note how the force and response are firstly “in sync” (  ~ 0 ). 39 Dr. the three phase angles are  2  0.10: Steady-state responses to illustrate phase angle.5 2 2. 52) to get: Dmax  1 2 1   2 (5.2.2.53) for   1. is very close to unity. C. This occurs when the denominator of equation (5.2. The corresponding maximum DAF is then given by substituting (5. as it should.Structural Analysis IV Chapter 5 – Structural Dynamics Maximum Steady-State Displacement The maximum steady-state displacement occurs when the DAF is a maximum. The other solution is:   1  2 2 (5.   0.54) Which for low values of damping.2.1 approximately.55) Which reduces to equation (5.2. Caprani .52) is a minimum: dD 0 d 12 2 d  1   2    2 2   0  d   2 2  1  4  1     4   2   0 2  2 2 2 1      2      1   2  2 2   0 The trivial solution to this equation of   0 corresponds to an applied forcing function that has zero frequency –the static loading effect of the forcing function.2. 40 Dr.54) into equation (5. this phase relationship allows the accurate measurements of the natural frequencies of structures.Structural Analysis IV Measurement of Natural Frequencies Chapter 5 – Structural Dynamics It may be seen from equation (5. That is. C.1 gave the natural frequency based on this type of test. 41 Dr.    2 .2. we change the input frequency  in small increments until we can identify a peak response: the value of  at the peak response is then the natural frequency of the system.50) that when   1. Caprani . Example 2. Caprani . C. we are only interested in the steady-state response. steady-sate and total response. We‟ve also used some of the equations from the Appendix to show the transient.2.Structural Analysis IV Chapter 5 – Structural Dynamics 5. 42 Dr. Normally however. which the total response approaches over time.8 Computer Implementation & Examples Using MS Excel Again we modify our previous spreadsheet and include the extra parameters related to forced response. Omega.the damping ratio theta = atan2((2*xi. m % v0 .^2).v0.xi).phase angle 43 Dr.mass.initial displacement. m/s % F . And another to return the phase angle (always in the region 0     ): function theta = phase(beta.the damping ratio D = 1.F./sqrt((1-beta. wd = w*sqrt(1-xi^2).u0. and modifying our previous damped response script.the frequency ratio % xi . w = sqrt(k/m). kg % k . theta = phase(beta.amplitude of forcing function.amplitude of vibration rad .k.^2)).xi) % This function returns the DAF. D. D = DAF(beta.stiffness.Structural Analysis IV Chapter 5 – Structural Dynamics Using Matlab First let‟s write a little function to return the DAF.initial velocity.length of time of required response % plotflag . rad/s % duration .the time vector at which the response was found % u .(1-beta.1 or 0: whether or not to plot the response % This function returns: % t . N/m % xi .xi. % refers to complex plane With these functions.xi) % This function returns the pahse angle.^2+(2*xi. we have: function [t u] = sdof_forced(m. % rad/s . N % Omega .*beta). C.*beta).frequency of forcing function. since we will use it often: function D = DAF(beta.damped circular frequency % % % % frequency ratio dynamic amplification factor m . associated with the parameters: % beta . Caprani . theta. associated with the % parameters: % beta . % compute the response at 1000 points delta_t = duration/(Npts-1).^2).plotflag) % This function returns the displacement of a damped SDOF system with % parameters: % m .circular natural frequency % rad/s .duration.the displacement vector of response Npts = 1000. ro = F/k*D. beta = Omega/w.xi).damping ratio % u0 .the frequency ratio % xi . if(plotflag == 1) plot(t. plot(t.1.'k--').*(Aconst*cos(wd*t)+Bconst*sin(wd*t)).15. plot(t. Caprani . end Running this for the same problem as before with F0  10 N and   15 rad/s gives: [t u] = sdof_forced(10.'Steady-State'). ylabel('Displacement (m)').100.Structural Analysis IV Chapter 5 – Structural Dynamics % Constants for the transient response Aconst = u0+ro*sin(theta).1). 44 Dr. t = 0:delta_t:duration.'Transient'. 0.6.025.0.20.'k').04 -0.u_transient. legend('Total Response'.u_steady.02 -0.u.04 Displacement (m) Total Response Transient Steady-State 0. xlabel('Time (s)'). the total response quickly approaches the steady-state response.0. C.02 0 -0. hold on.*t). u = u_transient + u_steady. u_steady = ro*sin(Omega*t-theta). Bconst = (v0+u0*xi*w-ro*(Omega*cos(theta)-xi*w*sin(theta)))/wd. u_transient = exp(-xi*w.06 0 1 2 3 Time (s) 4 5 6 As can be seen.'k:').06 0.0. hold off. xi(i)). end plot(beta.4.0001. xlabel('Frequency Ratio'). beta = 0.2.xi(i)).0].D).'Damping: 30%'.01:3. ylabel('Dynamic Amplification'). hold on.0.00001.. This gives: 6 5 Damping: Damping: Damping: Damping: Damping: Damping: Damping: Damping: Maxima 0% 10% 15% 20% 30% 40% 50% 100% Dynamic Amplification 4 3 2 1 0 0 0. 'Maxima').01:1. for i = 1:length(xi) D(i.99999. ylim([0 6]).15.1.1.'Damping: 15%'. C...0. % set y-axis limits since DAF at xi = 0 is enormous legend( 'Damping: 0%'. plot(betamax. Caprani . % very close to unity xi(1) = 0.'Damping: 10%'. end % A new xi vector for the maxima line xi = 0:0..0.'Damping: 40%'.54): % Script to plot DAF against Beta for different damping ratios xi = [0.'k--').0.:) = DAF(beta. Dmax(i) = DAF(betamax(i). 'Damping: 100%'. but this time showing the frequency ratio and maximum response from equation (5.0. xi(end) = 0.. 'Damping: 20%'.0.5 3 45 Dr.01:0.2.8.Dmax.5. 'Damping: 50%'.3.0. % very close to zero for i = 1:length(xi) betamax(i) = sqrt(1-2*xi(i)^2).5 Frequency Ratio 2 2.5 1 1..Structural Analysis IV Chapter 5 – Structural Dynamics Next let‟s use our little DAF function to plot something similar to Figure 2. grid on. xlabel('Frequency Ratio').T)..0001. 'Damping: 100%'..[0 45 90 135 180]). 46 Dr.1. using the phase function we wrote.0].0. beta = 0. legend('Damping: 0%'.01:0.9: % Script to plot phase against Beta for different damping ratios xi = [0.'ytick'.'Damping: 20%'.xi(i))*(180/pi).0. ylabel('Phase Angle (degrees)'). set(gca.Structural Analysis IV Chapter 5 – Structural Dynamics Lastly then. for i = 1:length(xi) T(i.'Damping: 10%'..1.01:3.'Damping: 50%'. C.'SE'). Caprani . ylim([0 180]).'Location'.2.0. we can generate Figure 2. % in degrees end plot(beta.5.:) = phase(beta. An assumed variation of acceleration.2. We will examine one method from the third category only.Structural Analysis IV Chapter 5 – Structural Dynamics 5. it is an important method and is extensible to non-linear systems. For arbitrary excitation we must resort to computational methods.56) There are three basic time-stepping approaches to the solution of the structural dynamics equations: 1. However. Caprani . Use of finite differences of velocity and acceleration. as well as multi degree-offreedom systems (MDOF). 2. which aim to solve the basic structural dynamics equation. at the next time-step: mui 1  cui 1  kui 1  Fi 1 (5. 47 Dr. C.9 Numerical Integration – Newmark’s Method Introduction The loading that can be applied to a structure is infinitely variable and closed-form mathematical solutions can only be achieved for a small number of cases. Interpolation of the excitation function.2. 3. Usual values are   1 1 1 and    .63) 48 Dr. C. Caprani .2.57) and (5.60) (5.58)) are sufficient to solve for the three unknown responses at each time step.Structural Analysis IV Chapter 5 – Structural Dynamics Development of Newmark’s Method In 1959 Newmark proposed a general assumed variation of acceleration method: ui 1  ui  1    t  ui   t  ui 1   2 2 ui 1  ui   t  ui   0.2.2.2.2.5    t   ui     t   ui 1     (5.59) (5. Newmark‟s equations can now be written as: (5.56).2. 6 2 The three equations presented thus far (equations (5. 2 4  Constant (average) acceleration is given by:    Linear variation of acceleration is given by:   1 1 and   .62) ui   t  ui   t  ui (5. For example: 6 4 2 1 1 and   . However to avoid iteration. we introduce the incremental form of the equations: ui  ui 1  ui ui  ui 1  ui ui  ui 1  ui Fi  Fi 1  Fi Thus.61) (5.2.2.57) (5.2. (5.2. t .58) The parameters  and  define how the acceleration is assumed over the time step. 65) and (5.2. Caprani .66) Next we use the incremental equation of motion.2.66) to get:  1 1 1  m ui  ui  2 ui    t  2     t         c  ui  ui  t 1    2    t     ui   k ui  Fi   (5.67) And introduce equations (5.2. derived from equation (5.2.2.68) Collecting terms gives: 49 Dr.56): m ui  c ui  k ui  Fi (5.Structural Analysis IV Chapter 5 – Structural Dynamics ui   t  ui  t   2 2 ui    t  ui 2 (5. C.2.63) and solving for the unknown increment in velocity gives: ui    t   ui      ui  t 1   ui   2  (5.64) Solving equation (5.64) for the unknown change in acceleration gives: ui  1 1 1 ui  ui 2 ui    t  2   t  (5.2.2.65) Substituting this into equation (5.2.2. 72) ˆ ˆ Since k and Fi are known from the system properties (m.2.2.2.65) respectively.71) Which are an effective stiffness and effective force at time i. t ). the algorithm properties (  .70)  1  1       ˆ Fi  Fi   m  c  ui   m  t   1  c  ui    2    2    t  (5. and the previous time-step ( ui . we can solve for the velocity and acceleration increments from equations (5.2. we can solve equation (5.2.69) ˆ k m (5.2. Caprani .73) Once the displacement increment is known. ui ). And once 50 Dr.2.2.66) and (5. k).  .69) becomes: ˆ ˆ k ui  Fi (5. Thus equation (5. c.2.72) for the displacement increment: ˆ Fi ˆ k ui  (5.Structural Analysis IV Chapter 5 – Structural Dynamics   1  m c  k  ui  2   t      t     1  1        Fi   m  c  ui   m  t   1  c  ui    2    2    t  Let‟s introduce the following for ease of representation: 1  k 2 c   t    t  (5. C. 2.74)    1 m  t   1 c 2  2  (5.2. Caprani .76) 51 Dr.2.59) to (5.2. The coefficients in equation (5. Newmark‟s method is stable if the time-steps is about t  0.2.71) become: ˆ Fi  Fi  Aui  Bui (5.1T of the system. so we can calculate these at the start as: A B 1  m c   t   (5. C.71) are constant (once t is). equations (5.2.75) Making equation (5.Structural Analysis IV Chapter 5 – Structural Dynamics all the increments are known we can compute the properties at the current time-step by just adding to the values at the previous time-step.61).2. 2.2.  and t .2. i. Calculate the coefficients for equation (5. c.80) (5. Initial calculations: a. Calculate the effective stiffness. k from equation (5.79) ˆ Fi ˆ k   t   ui      ui  t 1   ui   2  (5. 2.2.2.82) (5.2.74) and (5.2.77) ˆ b.71) from equations (5.2. Caprani .81) (5. C.Structural Analysis IV Chapter 5 – Structural Dynamics Newmark’s Algorithm 1.2. 3. Find the initial acceleration: u0  1  F0  cu0  ku0  m (5. Select algorithm parameters.83) (5.2.78) (5.84) ui  1 1 1 ui  ui 2 ui    t  2   t  ui  ui 1  ui ui  ui 1  ui ui  ui 1  ui 52 Dr. calculate: ˆ Fi  Fi  Aui  Bui ui  ui  (5. For each time step.2.2.70).75).  . Caprani . and see how the equations and algorithm are implemented. In the example shown. C.2.10 Computer Implementation & Examples Using MS Excel Based on our previous spreadsheet. Download it from the course website. we‟ve applied a sinusoidal load of 10 N for 0. we implement Newmark Integration.Structural Analysis IV Chapter 5 – Structural Dynamics 5.6 secs to the system we‟ve been using so far: 53 Dr. 1 or 0: whether or not to plot the response % The output is: % u . u(i+1) = u(i) + dui. ud(1) = ud0.circular natural frequency the damping coefficient % Calulate the effective stiffness keff = k + (gamma/(beta*dt))*c+(1/(beta*dt^2))*m. in equal time steps. N % u0 . m/s2 % Set the Newmark Integration parameters % gamma = 1/2 always % beta = 1/6 linear acceleration % beta = 1/4 average acceleration gamma = 1/2. m % ud . xi. N = length(t). dt = t(2)-t(1). damping ratio % t . m % v0 . initial displacement. C. k. displacement response.scalar. beta = 1/6. w = sqrt(k/m).scalar. N/m % xi .vector of length N. The input parameters are: % m . % calulate the change in force at each time step dF = diff(F). t. Bcoeff = (1/(2*beta))*m + dt*(gamma/(2*beta)-1)*c. % the initial acceleration for i = 1:(N-1) % N-1 since we already know solution at i = 1 dFeff = dF(i) + Acoeff*ud(i) + Bcoeff*udd(i).scalar. stiffness.vector of length N. We must write a completely new function that implements the Newmark Integration algorithm as we‟ve described it: function [u ud udd] = newmark_sdof(m. force at each time step. plotflag) % This function computes the response of a linear damped SDOF system % subject to an arbitrary excitation.vector of length N. % % % % the number of integration steps the time step rad/s .Structural Analysis IV Chapter 5 – Structural Dynamics Using Matlab There are no shortcuts to this one.vector of length N. u0. duddi = (1/(beta*dt^2))*dui-(1/(beta*dt))*ud(i)-(1/(2*beta))*udd(i). s % F . m/s % udd . c = 2*xi*k/w. dudi = (gamma/(beta*dt))*dui-(gamma/beta)*ud(i)+dt*(1gamma/(2*beta))*udd(i). % Set initial state u(1) = u0.scalar. ud0. Caprani . F. 54 Dr. m/s % plotflag . initial velocity. velocity response. acceleration response. udd(1) = (F(1)-c*ud0-k*u0)/m.vector of length N. dui = dFeff/keff.scalar. ud(i+1) = ud(i) + dudi. % Calulate the coefficients A and B Acoeff = (1/(beta*dt))*m+(gamma/beta)*c. mass. kg % k . i = i+1. C. Caprani .4) plot(t. with a huge range of applications. % set the time vector F = zeros(1. while t(i) < Tend F(i) = Famp*sin(pi*t(i)/Tend). end [u ud udd] = newmark_sdof(m.F. we must write a small script that sets the problem up and then calls the newmark_sdof function. ylabel('Acceleration (m/s2)').3) plot(t.0.6.Structural Analysis IV udd(i+1) = udd(i) + duddi. t. ud0 = 0. u0. end if(plotflag == 1) subplot(4.'k'). u0 = 0.1.1) plot(t.2) plot(t. The main difficulty is in generating the forcing function. k = 100. subplot(4. ylabel('Velocity (m/s)'). end Chapter 5 – Structural Dynamics Bear in mind that most of this script is either comments or plotting commands – Newmark Integration is a fast and small algorithm. xi = 0.udd.'k'). t = 0:0.u. % empty F vector % set sinusoidal force of 10 over 0. ud0. Tend = 0. xi.1. 1). F. subplot(4.1.1.ud. xlabel('Time (s)'). i = 1. xlabel('Time (s)'). ylabel('Displacement (m)'). In order to use this function. xlabel('Time (s)'). ylabel('Force (N)'). This produces the following plot: 55 Dr.length(t)).1:4. subplot(4.'k'). k. xlabel('Time (s)').6 s Famp = 10.'k').1. but it is not that hard: % script that calls Newmark Integration for sample problem m = 10. 5 1 1.5 1 1.1 0 0.5 3 3.5 4 Acceleration (m/s2) 1 0 -1 0 0.5 4 Displacement (m) 0.5 2 Time (s) 2.5 Velocity (m/s) 0 -0.5 1 1.Structural Analysis IV Chapter 5 – Structural Dynamics 10 Force (N) 5 0 -5 0 0.5 4 Explosions are often modelled as triangular loadings.5 1 1.5 3 3.5 2 Time (s) 2.5 2 Time (s) 2.5 2 Time (s) 2. Let‟s implement this for our system: 56 Dr.5 3 3. C.5 3 3.5 4 0.5 0 0.1 0 -0. Caprani . % N Tend = 0. ud0. 1). k = 100.0. Also notice that the acceleration response is the most sensitive – this is the most damaging to the building. t. 57 Dr. Caprani . i = i+1. while t(i) < Tend F(i) = Fmax*(1-t(i)/Tend). % s t = 0:0. as force is mass times acceleration: the structure thus undergoes massive forces.Structural Analysis IV Chapter 5 – Structural Dynamics % script that finds explosion response m = 10. possibly leading to damage or failure. Fmax = 50. F. u0 = 0. u0. the vibrations will take several periods to dampen out. As can be seen from the following plot. end [u ud udd] = newmark_sdof(m.1. % empty F vector % set reducing triangular force i = 1. ud0 = 0. even though the explosion only lasts for a brief period of time.2. C. xi = 0. % set the time vector F = zeros(1. xi.length(t)). k.01:2. Structural Analysis IV Chapter 5 – Structural Dynamics 60 Force (N) 40 20 0 0 0.5 3 3.5 1 1.5 3 3. C.5 1 1. Caprani .5 1 1.5 2 Time (s) 2.5 0 -0.5 4 Velocity (m/s) 0.5 4 58 Dr.5 0 0.5 3 3.5 2 Time (s) 2.5 4 Displacement (m) 0.5 2 Time (s) 2.5 2 Time (s) 2.2 0 -0.2 0 0.5 4 Acceleration (m/s2) 5 0 -5 0 0.5 3 3.5 1 1. 44 mm was noted.223 sec.5×103 kNm2.222 sec. 100 kN Problem 3 An SDOF system (m = 20 kg. what deflection will result if a 50 kN horizontal load is applied? You may neglect the mass of the tower.2. Ans: 0. Given that the mass of the tank is 150 tonnes. 8 m wide single-bay single-storey frame is rigidly jointed with a beam of mass 5.91 mm. Find the force required to deflect the frame 25 mm laterally. Ans: 4. Caprani . 3. C.502 sec. The period of motion was measured at 0. On the first return swing a deflection of 19.502 Hz. 1.Structural Analysis IV Chapter 5 – Structural Dynamics 5. (a) Find the natural frequency.2 mm Problem 2 A 3 m high. (b) the period of vibration.666 Hz. Assuming that the stiffness of the columns cannot 59 Dr. k = 350 N/m) is given an initial displacement of 10 mm and initial velocity of 100 mm/s. Ans: 50. 25.11 Problems Problem 1 A harmonic oscillation test gave the natural frequency of a water tower to be 0.285 sec. and (d) the time at which the third maximum peak occurs. a jack applied a load of 100 kN and then instantaneously released. Calculate the natural frequency in lateral vibration and its period. (c) the amplitude of vibration.41 Hz.000 kg and columns of negligible mass and stiffness of EIc = 4. 0. Problem 4 For the frame of Problem 2. Caprani . Problem 5 From the response time-history of an SDOF system given: 60 Dr.367 kg·s/m. and (d) the amplitude after 5 cycles.04. 11.2228 sec. Ans: 0. (b) the coefficient of damping. 0. 7. (c) the undamped frequency and period. C. find (a) the damping ratio. 4.Structural Analysis IV Chapter 5 – Structural Dynamics change.11 mm.488 Hz. 600 kg is place in the middle (point C) of an 8 m long simply supported beam (EI = 8×103 kNm2) of negligible mass.9 sec. An engineer investigated and found the natural period in sway to be 0. determine the steady-state amplitude and deflection at C.75 Hz.34 mm. Problem 6 Workers‟ movements on a platform (8 × 6 m high. 4.021 Hz. 0. Ans: 4.41 mm. C. (b) use the half amplitude method to calculate the damping ratio. 206.92 mm. 0.4 has a reciprocating machine put on it.026 Hz. 22.66 mm.56 mm. 36. and (c) calculate the undamped natural frequency and period. 26. (a) What is the steadystate amplitude of vibration if the damping ratio is 4%? (b) What would the steadystate amplitude be if the forcing frequency was in resonance with the structure? Ans: 2. Caprani . The mass of this machine is 4 tonnes and is in addition to the mass of the beam.3 sec. Problem 7 The frame of examples 2. m = 200 kN) are causing large dynamic motions. What tie diameter is required? Ans: 28. Diagonal remedial ties (E = 200 kN/mm2) are to be installed to reduce the natural period to 0.248 sec.05.2 and 2. The machine exerts a periodic force of 8. Problem 8 An air conditioning unit of mass 1. The motor runs at 300 rpm and produces an unbalanced load of 120 kg. What rpm will result in resonance and what is the associated deflection? Ans: 1. Assuming a damping ratio of 5%.7 rpm.5 kN at a frequency of 1.Structural Analysis IV Chapter 5 – Structural Dynamics (a) estimate the damped natural frequency.1 mm. 61 Dr. 3.1 s duration and magnitude 10 N at time 1.015 0.8 m/s.5 1 1.1 s.75 MN/m.7 m/s2 Problem 11 Determine the maximum response of a system (m = 1. 14.5 MN/m and no damping.005 0 -0.  = 10%) when subjected to an increasing triangular load which reaches 22.6 mm.01 0. k = 1. Ans. 120 mm. Ans.39 m/s.025 0. Caprani . stiffness 17.015 0 0. Do this up for a duration of 4 s.5 3 3.5 4 Problem 10 Determine the maximum responses of a water tower which is subjected to a sinusoidal force of amplitude 445 kN and frequency 30 rad/s over 0.5 2 Time (s) 2. Ans. 15.2 kN after 0. when acted upon by an impulse of 0.05 m/s. with initial velocity of 0. The tower has properties.0 s.5 t. C.005 -0.01 -0.02 Displacement (m) 0. 0.75 t.Structural Analysis IV Chapter 5 – Structural Dynamics Problem 9 Determine the response of our example system. below 0.3 secs.0 m/s2 62 Dr. 120. mass 17. 3. we can see that the forces that act on the masses are similar to those of the SDOF system but for the fact that the springs. forces and deflections may all differ in properties. from the same figure.Structural Analysis IV Chapter 5 – Structural Dynamics 5. (b) and (c) Free-body diagrams of forces Considering Figure 3.1: (a) 2DOF system.1 General Case (based on 2DOF) (a) (b) (c) Figure 3. masses. we can see the interaction forces between the masses will result from the relative deflection between the masses. Also. Caprani . F x  0 .1. For each mass. hence: m1u1  c1u1  k1u1  c2  u1  u2   k2  u1  u2   F1 (5.3.1) 63 Dr. the change in distance between them. C. dashpots.3 Multi-Degree-of-Freedom Systems 5. is the vector of velocity for each DOF.3.4) Or another way: Mu + Cu + Ku = F (5.3.3.Structural Analysis IV Chapter 5 – Structural Dynamics m2u2  c2  u2  u1   k2 u2  u1   F2 (5. is the vector of displacements for each DOF. is the load vector.3.3) This can be written in matrix form:  m1 0  0   u1  c1  c2   m2  u2   c2   c2   u1   k1  k2   c2  u2   k2   k2   u1   F1     k2  u2   F2   (5.3. is the damping matrix (symmetrical matrix). is the vector of the accelerations for each DOF. Re-arranging we get: u1  c1  c2  u2  c2  u2  c2  u1  k1  k2  u2  k2  u1  k2  u 2  k 2   F1  F2 u1m1 u2 m2 u1  c2  (5. is the stiffness matrix (symmetrical matrix).5) is quite general and reduces to many forms of analysis: 64 Dr. K u F Equation (5. Caprani . C.2) In which we have dropped the time function indicators and allowed u and u to absorb the directions of the interaction forces.5) where: M u C u is the mass matrix (diagonal matrix). 3. C.3.9) We will restrict our attention to the case of undamped free-vibration – equation (5. 65 Dr.3.7) Undamped forced vibration: Mu + Ku = F (5.3.as the inclusion of damping requires an increase in mathematical complexity which would distract from our purpose.3.6) Undamped free vibration: Mu + Ku = 0 (5. Caprani .8) Static analysis: Ku = F (5.Structural Analysis IV Chapter 5 – Structural Dynamics Free vibration: Mu + Cu + Ku = 0 (5.7) . 14) For the 2DOF system.3.10) where a is the vector of amplitudes corresponding to each degree of freedom. From this we get: u =  2a sin t      2u (5. substitution of (5.3.12) Since the sine term is constant for each term: K   2M  a = 0   (5. hence. we have: 66 Dr. in order for (5. a  0 in general.3.3.3. Caprani .3. so following that method (equation (2.10) and (5.3. In addition we see that the problem is a standard eigenvalues problem.3.3. C.3.42)).3. by Cramer‟s rule.13) to hold the determinant of K   2M must then be zero: K   2M = 0 (5. we propose a solution of the form: u = a sin t    (5.2 Free-Undamped Vibration of 2DOF Systems The solution to (5.7) yields:  2Ma sin t    + Ka sin t    = 0 (5.Structural Analysis IV Chapter 5 – Structural Dynamics 5.13) We note that in a dynamics problem the amplitudes of each DOF will be non-zero.11) into (5. Hence.11) Then.7) follows the same methodology as for the SDOF case. normally.15) Expansion of (5. this means there are two values of  2 ( 12 and 22 ) that will satisfy the relationship. For each n2 substituted back into (5. some modes may be critical design cases depending on the type of harmonic loading as will be seen later. relative to. In our case.d). thus there are two frequencies for this system (the lowest will be called the fundamental frequency). we will get a certain amplitude vector a n .15) leads to an equation in  2 called the characteristic polynomial of the system. not just in the fundamental mode. we will not know the absolute values of the amplitudes as it is a free-vibration problem. As we will see in the following example. the first value in a n . but also in higher harmonics. There will be two solutions or roots of the characteristic polynomial in this case and an n-DOF system has n solutions to its characteristic polynomial. From our analysis of SDOF systems it‟s apparent that should any loading coincide with any of these harmonics.3. Caprani .3. However. φ n .Structural Analysis IV Chapter 5 – Structural Dynamics K   2M =  k2  k1    2m1  k2   2m2   k22  0    (5.13). The solutions of  2 to this equation are the eigenvalues of K   2M    . large DAF‟s will result (Section 2. This means that each frequency will have its own characteristic displaced shape of the degrees of freedoms called the mode shape. Thus. C.3. 67 Dr. the implication of the above is that MDOF systems vibrate. hence we express the mode shapes as a vector of relative amplitudes. 3.2) has very stiff floor slabs relative to the supporting columns. or shear stiffness of the columns is: 12 EI  k1  k2  k  2  3 c   h  2  12  4.5  106 k  33  4  106 N/m The characteristic polynomial is as given in (5. Calculate the natural frequencies and mode shapes.3: 2DOF model of the shear frame.2: Shear frame problem.Structural Analysis IV Chapter 5 – Structural Dynamics 5.5  103 kNm2 Figure 3. The lateral.3. C. Caprani .15) so we have: 68 Dr. Figure 3.3 Example of a 2DOF System The two-storey building shown (Figure 3.3. EI c  4. We will consider the free lateral vibrations of the two-storey shear frame idealised as in Figure 3. 6  ω n 3.4  10   16  10  0 Chapter 5 – Structural Dynamics This is a quadratic equation in  2 and so can be solved using a  15 106 .7241  4 69 Dr. First solve for the E  K   2M      matrix and then solve Ea = 0 for the amplitudes a n .8735 6 E1     10 2. This may be written: 425. for a 2DOF system.13): K   2M  a = 0 .28   ω2    rad/s and f   Hz  hence ω n   n 2 7.Structural Analysis IV 8  106   2 5000  4  106   2 3000  16  1012  0    6 4 10 2 12 15  10   4. Then.97   50. b  4. form φ n . equation (5. C.3 20. In general.1  2508  To solve for the mode shapes.3 and 22  2508 .4 1010 and c  16 1012 in the usual expression 2  b  b 2  4ac 2a Hence we get 12  425.3. we have: k  k En   1 2   k2 k2  m  n2  1 k2   0 0   k1  k2  n2 m1  m2    k2   k2   k2  n2 m2  For 12  425. Caprani . we will use the appropriate form of the equation of motion.3 : 4  5. 1  425.135  2508  For our frame. we calculate: 5.881a2  0.8735a1  4a2  0  a1  0.4.681a2   1   φ1   1  4a1  2. 70 Dr.524a2  0  a1  0.681  Similarly for 22  2508 : 4   4.7241 a2  0  4 Taking either equation.7241a2  0  a1  0. C.8735 E1a1  106       2.881a2   1  φ2   1  4a1  3.468 1. again taking either equation.524   4 Hence.681a2  0.Structural Analysis IV Chapter 5 – Structural Dynamics Hence 4   a1  0 5. Caprani . we calculate: 4. we can sketch these two frequencies and associated mode shapes: Figure 3.881  The complete solution may be given by the following two matrices which are used in further analysis for more complicated systems.54a1  4a2  0  a1  0.54 6 E2     10 3.3  1 and Φ   ω2    n  1. 5L and L as well as half the cantilever mass „lumped‟ at the midpoint and one quarter of it lumped at the tip. The mode shapes are shown in Figure 3. There are different mode shapes for different forms of deformation. consider a 2DOF idealisation of a cantilever which assumes stiffness proportional to the static deflection at 0. with more DOFs it will approach a better solution. 71 Dr. As an example. but. C. higher harmonics may also be important.Structural Analysis IV Chapter 5 – Structural Dynamics Figure 3.4: Mode shapes and frequencies of the example frame. lateral and vertical for example. torsional. Caprani . Periodic loads acting in these directions need to be checked against the fundamental frequency for the type of deformation.5. Larger and more complex structures will have many degrees of freedom and hence many natural frequencies and mode shapes. In Section 4(a) we will see the exact mode shape for this – it is clear that the approximation is rough. 5: Lumped mass. Caprani . 2DOF idealisation of a cantilever. 72 Dr.Structural Analysis IV Chapter 5 – Structural Dynamics Mode 1 Mode 2 Figure 3. C. 17 4. The type of 73 Dr.07 +16% +7% +4% +5% +11% +3% +4% +2% 20% +1% -1% +21% +0% +19% Figure 3.20 5.63 4.50 3.91 3.Structural Analysis IV Chapter 5 – Structural Dynamics 5. This is taken from a paper by Dr.49 2.98 1.04 3.00 4.29 5.11 3.14 1.6: Undeformed shape 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Table 1: Modal frequencies Table 1 gives the first 14 mode and associated frequencies from both direct measurements of the bridge and from finite-element modelling of it.72 6.73 3. we will look at the results of some research conducted into the behaviour of this bridge which forms part of the current research into lateral synchronise excitation discovered on the London Millennium footbridge. Mode Mode Type L1 V1 V2 V3 L2 V4 V5 V6 T1 V7 V8 T2 L3 T3 Measured Frequency (Hz) 0.93 4.72 Predicted Frequency (Hz) 1. Caprani .72 5.48 4. Scotland Returning to the case study in Section 1.3.90 5.62 3.45 4.86 2.4 Case Study – Aberfeldy Footbridge.94 2.01 3. Paul Archbold.52 1.40 4.63 1. formerly of University College Dublin. C. the modes have been estimated in the correct sequence and there may be some measurement error. however. Also note that the overall fundamental mode is lateral – this was the reason that this bridge has been analysed – it is similar to the Millennium footbridge in this respect. Caprani .7. there are a lot of modes that may be excited by this loading.Structural Analysis IV Chapter 5 – Structural Dynamics mode is also listed. V is vertical and T is torsional.7 illustrates the dynamic motion due to a person walking on this bridge – this is probably caused by the third or fourth mode. We can see now that (from Section 1) as a person walks at about 2. 74 Dr. L is lateral.8 Hz. Figure 1. It can be seen that the predicted frequencies differ slightly from the measured. Several pertinent mode shapes are given in Figure 3. C. 94 Hz Mode 9: 1st Torsional mode 4.14 Hz Mode 2: 1st Vertical mode 1. Caprani . 75 Dr.63 Hz Mode 3: 2nd Vertical mode 1.7: Various Modes of Aberfeldy footbridge. C.Structural Analysis IV Chapter 5 – Structural Dynamics Mode 1: 1st Lateral mode 1.17 Hz Figure 3. C. For some basic structures 76 Dr. the more DOF‟s used the more accurate the model (Section 3. Discretization into an MDOF structure is certainly an option and is the basis for finite-element dynamic analyses. as with any continuous structure. (b) resultant forces acting on the differential element. any continuous structure has an infinite number of degrees of freedom. Caprani . it may be seen that any differential element will have an associated stiffness and deflection – which changes with time – and hence a different acceleration.4.1.Structural Analysis IV Chapter 5 – Structural Dynamics 5.1 Exact Analysis for Beams General Equation of Motion Figure 4.1: Basic beam subjected to dynamic loading: (a) beam properties and coordinates.b).4 Continuous Structures 5. Thus. In examining Figure 4. assuming constant mass):  2 v  x. A general expression can be derived and from this. t  t 2  p  x.1(b). the exact behaviour can be explicitly calculated. t  dx  0 (5. We will limit ourselves to free-undamped vibration of beams that are thin in comparison to their length. C. hence: p  x.4.2) Thus we have. Caprani . several usual cases may be established. t   m (5.3) 77 Dr.1) after having cancelled the common V  x. Consider the element A of Figure 4. t  dx  V  x.Structural Analysis IV Chapter 5 – Structural Dynamics though. t  t 2 f I  x. Figure 4. after dividing by the common dx term: V  x. F y  0 .4. The resultant transverse inertial force is (mass × acceleration.4.2: Instantaneous dynamic deflected position. t  x dx  f I  x. t  shear term. t  dx  mdx (5. t  x  2 v  x. t   (5.4. t  t 2 EI m 0 (5. and dropping second order and common terms.3). with no acceleration. t  x 4  2 v  x. Caprani .4.4) Differentiating this with respect to x and substituting into (5. is the usual static relationship between shear force and applied load. t  t 2  p  x. C. in addition to the 2 relationship M  EI  v x 2 (which assumes that the beam is of constant stiffness): EI  4v  x.Structural Analysis IV Chapter 5 – Structural Dynamics which. By taking moments about the point A on the element. we get the usual expression: M  x.4.5) With free vibration this is:  4 v  x.6) 78 Dr. t  x V  x. t  (5.4. t  x 4 m  2 v  x.  It is a second order differential in t and so we will need two temporal initial conditions to solve – initial deflection and velocity at a point for example.4. assume the solution is of a form of separated variables: v  x.6) yields several aspects:  It is separated into spatial ( x ) and temporal ( t ) terms and we may assume that the solution is also. C.4.10) Y  t    2Y  t   0 79 Dr.6) and collecting terms we have: 4 2 EI 1    x  1  Y t    constant   2 4 2 m   x  x Y  t  t (5. Hence.Structural Analysis IV Chapter 5 – Structural Dynamics General Solution for Free-Undamped Vibration Examination of equation (5. each function type (spatial or temporal) is equal to  2 and so we have:  4  x  x 4 EI   2 m  x  (5.4. hence we will need four spatial boundary conditions to solve – these will come from the support conditions at each end. To begin. t     x  Y  t  (5.8) This follows as the terms each side of the equals are functions of x and t separately and so must be constant.9) (5.7) where   x  will define the deformed shape of the beam and Y  t  the amplitude of vibration. Caprani .4.4.4.  It is a fourth-order differential in x . Inserting the assumed solution into (5. 13) There are then four roots for s and when each is put into (5.4.4. sin.14) In which the G ‟s may be complex constant numbers. substitution into (5.12) And assuming a solution of the form   x   G exp(sx) . C.11) In order to evaluate  we will use equation (5.Structural Analysis IV Chapter 5 – Structural Dynamics Equation (5. the fourth however is arbitrary and will depend on  . three of which may be evaluated through the boundary conditions.17)): Y  Y  t   Y0 cos  t   0  sin  t   (5.4)) and so the solution must be of the same form (equation (2.4.4. by using Euler‟s expressions for cos. but.4.13) and added we get:   x   G1 exp  i x   G2 exp  i x   G3 exp  x   G4 exp   x  (5.4.15) where the A ‟s are now real constants. Caprani . sinh and cosh we get:   x   A1 sin  x   A2 cos  x   A3 sinh  x   A4 cosh   x  (5.4. 80 Dr.9) and we introduce: 4   2m EI (5.4.10) is the same as for an SDOF system (equation (2.9) gives: s 4   4  G exp  sx   0 (5.4. 3: First three mode shapes and frequency parameters for an s-s beam.18) from which. Caprani . C. Similarly.4. t   0 and EI 2  0.14) we find A2  A4  0 .4.16) into equation (5. t   0 and EI 2  L.17) Substituting (5.4.17) gives:   L   A1 sin( L)  A3 sinh( L)  0  ''  L    2 A1 sin( L)   2 A3 sinh( L)  0 (5. we get two possibilities: 81 Dr. The boundary conditions consist of zero deflection and bending moment at each end:  2v v  0. t   0 x (5. t   0 x  2v v  L.4. (5.Structural Analysis IV Chapter 5 – Structural Dynamics Simply-supported Beam Figure 4.4.4.16) (5. 21) and the corresponding modes shapes are therefore: n  x   A1 sin   n x    L  (5. Hence.3.4. The first three mode shapes and frequencies are shown in Figure 4.20) which is the frequency equation and is only satisfied when  L  n . from (5.4.4. and so the non-trivial solution A1  0 must give us: sin( L)  0 (5.19) however. C. 82 Dr. Caprani . since sinh( x) is never zero.4.4.Structural Analysis IV Chapter 5 – Structural Dynamics 0  2 A3 sinh( L) 0  A1 sin( L) (5.22) where A1 is arbitrary and normally taken to be unity. We can see that there are an infinite number of frequencies and mode shapes ( n   ) as we would expect from an infinite number of DOFs. A3 must be.12) we get:  n  n     L  2 EI m (5. 4.27) 83 Dr.23) (5.4.25) where a prime indicates a derivate of x .14) we get A4   A2 and A3   A1 . (5.4.24)  2v  3v L.4.4. t   0 and EI 3  L. C.4. Substituting (5.24) gives:  ''  L    2 A1 sin( L)   2 A2 cos( L)   2 A3 sinh( L)   2 A4 cosh( L)  0  '''  L    3 A1 cos( L)   3 A2 sin( L)   3 A3 cosh( L)   3 A4 sinh( L)  0 (5.26) Solving for A1 and A2 we find:  cos( L)  cosh( L) 2  A1  0   sin( L)  sinh( L)   sin( L)  sinh( L)      2  cos( L)  cosh( L)   A2  0   sin( L)  sinh( L)   sin( L)  sinh( L)      (5. Similarly. and so we find: A1  sin( L)  sinh( L)   A2  cos( L)  cosh( L)   0 A1  cos( L)  cosh( L)   A2   sin( L)  sinh( L)   0 (5. The boundary conditions consist of: v  0.4.4. Caprani . t   0  x 2 x Which represent zero displacement and slope at the support and zero bending moment and shear at the tip.Structural Analysis IV Chapter 5 – Structural Dynamics Cantilever Beam This example is important as it describes the sway behaviour of tall buildings. t   0 x (5.23) into equation (5. t   0 and EI v  0. Structural Analysis IV Chapter 5 – Structural Dynamics In order that neither A1 and A2 are zero, the expression in the brackets must be zero and we are left with the frequency equation: cos( L)cosh( L)  1  0 (5.4.28) The mode shape is got by expressing A2 in terms of A1 : A2   sin( L)  sinh( L) A1 cos( L)  cosh( L) (5.4.29) and the modes shapes are therefore: sin( x)  sinh( x)    n  x   A1 sin( L)  sinh( L)    cosh( x)  cos( x)  cos( L)  cosh( L)     (5.4.30) where again A1 is arbitrary and normally taken to be unity. We can see from (5.4.28) that it must be solved numerically for the corresponding values of  L The natural frequencies are then got from (5.4.21) with the substitution of  L for n . The first three mode shapes and frequencies are shown in Figure 4.4. 84 Dr. C. Caprani Structural Analysis IV Chapter 5 – Structural Dynamics Figure 4.4: First three mode shapes and frequency parameters for a cantilever. 85 Dr. C. Caprani Structural Analysis IV Chapter 5 – Structural Dynamics 5.4.2 Approximate Analysis – Bolton’s Method We will now look at a simplified method that requires an understanding of dynamic behaviour but is very easy to implement. The idea is to represent, through various manipulations of mass and stiffness, any complex structure as a single SDOF system which is easily solved via an implementation of equation (1.2): f  1 2 KE ME (5.4.31) in which we have equivalent SDOF stiffness and mass terms. Consider a mass-less cantilever which carries two different masses, Figure 4.5: Figure 4.5: Equivalent dynamic mass distribution for a cantilever. The end deflection of a cantilever loaded at its end by a force P is well known to be PL3 3EI and hence the stiffness is 3EI L3 . Therefore, the frequencies of the two cantilevers of Figure 4.5 are: f1  1 2 3EI M 1 x3 (5.4.32) fE  1 2 3EI ; M E L3 (5.4.33) 86 Dr. C. Caprani Figure 4. we have an expression for the deflection at a point: 87 Dr.Structural Analysis IV Chapter 5 – Structural Dynamics And so. if the two frequencies are to be equal. Caprani . the corresponding part of M E is: x dM E    mdx L 3 (5.4. it ignores the fact that every element affects the deflection (and hence vibration) of every other element.35) Therefore the cantilever with self-mass uniformly distributed along its length vibrates at the same frequency as would the mass-less cantilever loaded with a mass one quarter its actual mass. The answer is reasonable for design though.6: Equivalent dynamic mass distribution for an s-s beam Similarly for a simply supported beam.4.34) and integrating: x M E     mdx 0 L  0. C.25mL L 3 (5. This answer is not quite correct but is within 5%. and considering M 1 as the mass of a small element dx when the mass per metre is m . Proceeding in a similar way we can find equivalent spring stiffnesses and masses for usual forms of beams as given in Table 1. Table 4.1 however.6. also includes a refinement of the equivalent masses based on the known dynamic deflected shape rather than the static deflected shape.37) Considering Figure 4.4. from (5.4. Caprani .Structural Analysis IV Chapter 5 – Structural Dynamics Px 2  L  x  x  3EIL and so its stiffness is: 2 (5. C.39)  8/15mL which is about half of the self-mass as we might have guessed.31): 3EIL 48EI  3 2 x2  L  x  M1 L M E (5.4.4. 88 Dr.4. we see that.36) Kx  3EIL 2 x2  L  x  (5.38) and as the two frequencies are to be equal: M E  16 x 0 L 2  L  x L4 2 mdx (5. 7: Effective SDOFs: (a) neglecting relative amplitude. Figure 4. C. the continuity over the supports requires all the spans to vibrate at the same frequency for each of its modes. In considering continuous beams. stiffnesses and relative amplitudes. Thus we may consider 89 Dr.1: Bolton‟s table for equivalent mass. (b) including relative amplitude. Caprani .Structural Analysis IV Chapter 5 – Structural Dynamics Table 4. 90 Dr.72. It is equivalent to the SDOF model of Figure 4.8: Continuous beam of Examples 1 to 3.45. we would have the model of Figure 4. 40.7(b) which would be more accurate – especially when there is a significant difference in the member stiffnesses and masses: long heavy members will have larger amplitudes than short stiff light members due to the amount of kinetic energy stored.30. the stiffness and mass of each span must be weighted by its relative amplitude before summing. if we allowed for the relative amplitude between the different spans. Figure 4.53 for the first eight modes.10. 13. 46. But.7(a). C.Structural Analysis IV Chapter 5 – Structural Dynamics summing the equivalent masses and stiffnesses for each span and this is not a bad approximation.67. the exact multipliers are known to be 10. Caprani . Consider the following examples of the beam shown in Figure 4. Thus.32. 17. 53. 21.8.89 and 60. Structural Analysis IV Chapter 5 – Structural Dynamics Example 1: Ignoring relative amplitude and refined ME From Table 4. Caprani . we have: K E  3 48EI 101.60  1 2 EI mL4 The multiplier in the exact answer is 10.9  .1 and the previous discussion.30: an error of 5%. and L3 M E 8 1   mL  3    .655mL and applying (5.4928mL 1  0. 91 Dr.4108  1.31) we have: f  10.9%.4. and the previous discussion: K E  EI  48  3  101.9 EI EI 1   0.82  1 2 EI mL4 The multiplier in the exact answer is 10.9 3 3 3 L L L M E  3  0.31) we have: f  10.30: a reduced error of 2.4. Example 2: Including relative amplitude and refined ME From Table 4.4108  185.1. C.  15 2  and applying (5.4299mL  0. 9: Assumed mode shape for which the frequency will be found. we have: f   40.8 1 2 EI mL4 The multiplier in the exact answer is 40. C.9%.5L  EI L3  M E  7  0.9 EI 1  3 3  0.5L   0.7.31). The mode shape for calculation is shown in Figure 4.1 we have: K E  7 48EI 101.4108  3022.9  1. Hence we have seven simply supported half-spans and one cantilever half-span.4.813mL again. applying (5.4928m  0. 92 Dr.5L   1  0.45: and error of 0. We can assume supports at the midpoints of each span as they do not displace in this mode shape.Structural Analysis IV Chapter 5 – Structural Dynamics Example 3: Calculating the frequency of a higher mode Figure 4.4108  0. so from Table 4.4299m  0. Caprani .5L   0. lower modes have a larger R and higher modes have a smaller R . C. so for any structure we can say that when it is vibrating at its natural frequency it has very low stiffness – and in the case of no damping: zero stiffness. Figures 4. Higher modes will have higher stiffnesses but stiffness may also be recognised in one form as M 1  EI R (5.d described how the DAF is very large when a force is applied at the natural frequency of the structure.e.Structural Analysis IV Chapter 5 – Structural Dynamics Mode Shapes and Frequencies Section 2. smaller stiffnesses have a larger R and larger stiffnesses have a smaller R . Noting that a member in single curvature (i.3 and 4.4 illustrate this clearly. Caprani . Therefore. no point of contraflexure) has a larger R than a member in double curvature (1 point of contraflexure) which in turn has a larger R than a member in triple curvature (2 points of contraflexure). Similarly then. we can distinguish modes by deflected shapes.40) where R is the radius of curvature and M is bending moment. This enables us to distinguish between modes by their frequencies.4. 93 Dr. symmetry may exist about a support and so reduced structures may be used to estimate the frequencies of the total structure. 94 Dr. for non-identical spans. An important fact may be deduced from Figure 4. Also.10: Typical modes and reduced structures.10 and the preceding arguments: a continuous beam of any number of identical spans has the same fundamental frequency as that of one simply supported span: symmetrical frequencies are similarly linked. Caprani .Structural Analysis IV Chapter 5 – Structural Dynamics Figure 4. C.10(b) and (d) for symmetrical and antisymmetrical modes. reductions are shown in Figure 4. Ans.5×103 kNm2 and m = 100 kg/m? Ans. 95 Dr. Caprani . fixed at A and on rollers at B and C.: 9. It is known that a load of 10 kN causes a 3 mm deflection. Problem 2 Calculate the first natural frequency of a 4 m long cantilever (EI = 4. Ans.76 Hz.95Hz.4.5m.: 6.: 3. Span AB is 8 m with flexural stiffness of 2EI and a mass of 1.320 kNm2) which carries a mass of 500 kg at its centre and has self weight of 1200 kg.: 3.74 Hz. ? Hz.3 Hz. Problem 3 What is the fundamental frequency of a 3-span continuous beam of spans 4. Problem 4 Calculate the first and second natural frequencies of a two-span continuous beam. 8 and 5 m with constant EI and m? What is the frequency when EI = 6×103 kNm2 and m = 150 kg/m? Ans. Span BC is 6 m with flexural stiffness EI and mass m per metre.Structural Analysis IV Chapter 5 – Structural Dynamics 5. C. What are the frequencies when EI = 4.3 Problems Problem 1 Calculate the first natural frequency of a simply supported bridge of mass 7 tonnes with a 3 tonne lorry at its quarter point. 96 Dr. 5. C.: ? Hz. 4 and 5 m with constant EI and m? What are the frequencies when EI = 4×103 kNm2 and m = 120 kg/m? What are the new frequencies when support A is fixed? Does this make it more or less susceptible to human-induced vibration? Ans. Caprani .Structural Analysis IV Problem 5 Chapter 5 – Structural Dynamics Calculate the first and second natural frequencies of a 4-span continuous beam of spans 4. ? Hz. 5 Practical Design Considerations 5. It has 97 Dr. C.Structural Analysis IV Chapter 5 – Structural Dynamics 5.1 Human Response to Dynamic Excitation Figure 5.5. Caprani .1: Equal sensation contours for vertical vibration The response of humans to vibrations is a complex phenomenon involving the variables of the vibrations being experienced as well as the perception of it. This graph shows that for a given frequency. as the amplitude gets larger it becomes more uncomfortable.Structural Analysis IV Chapter 5 – Structural Dynamics been found that the frequency range between 2 and 30 Hz is particularly uncomfortable because of resonance with major body parts (Figure 5. thus it is acceleration that is governing the comfort. This may also be realised from car-travel: at constant velocity nothing is perceptible. Other factors are also important: the duration of exposure. Caprani . the difference has the effect of moving the illustrated bands up a level. This is important in the design of tall buildings which sway due to wind loading: it is the acceleration that causes discomfort. upon rapid acceleration the motion if perceived ( F  ma ). waveform (which is again linked to acceleration). C. and. 98 Dr.1. type of activity.1 have been obtained for each direction of vibration but vertical motion is more uncomfortable for standing subjects. Sensation contours for vertical vibrations are shown in Figure 5. for the transverse and longitudinal cases.2). but. Figure 5. psychological factors. An example is that low frequency exposure can result in motion sickness.2: Human body response to vibration Response graphs like Figure 5. Structural Analysis IV Chapter 5 – Structural Dynamics 5. the peak acceleration is got from: a0   0. Caprani . Vibration limits for light floors from the 1984 Canadian Standard is shown in Figure 5.5.9  2 f I M (5.2 Crowd/Pedestrian Dynamic Loading Lightweight Floors Figure 5. C.2.1) where I is the impulse (the area under the force time graph) and is about 70 Ns and M is the equivalent mass of the floor which is about 40% of the distributed mass.3: Recommended vibration limits for light floors.5. 99 Dr. Structural Analysis IV Chapter 5 – Structural Dynamics This form of approach is to be complemented by a simple analysis of an equivalent SDOF system. human loading should not be problematic. by keeping the fundamental frequency above 5 Hz. 100 Dr. Also. Caprani . as seen in Section 1. C. Once again.5 Hz. Caprani .Structural Analysis IV Chapter 5 – Structural Dynamics Crowd Loading This form of loading occurs in grandstands and similar structures where a large number of people are densely packed and will be responding to the same stimulus.3 seconds – the method for lightweight floors can be applied to this scenario. however. allowance should also be made due to the crowd-sway that may accompany some events a value of about 0. In the transverse or longitudinal directions.97 at about 2. for example. Staircases can be subject to considerable dynamic forces as running up or down such may cause peak loads of up to 4-5 times the persons bodyweight over a period of about 0. normally generates frequencies of 2 – 3 Hz. by keeping the natural frequency of the structure above about 5 Hz no undue dynamic effects should be noticed. Coordinated jumping to the beat of music.3 kN per metre of seating parallel and 0. can cause a DAF of about 1.15 kN perpendicular to the seating is an approximate method for design. 101 Dr. Dancing. C. the lessons of the London Millennium and the Tacoma Narrows bridges need to be heeded though: dynamic effects may occur in any direction or mode that can be excited by any form of loading.1 General.4. Caprani .30) and (3. An approximate method for checking foot bridges is the following: umax  ust K (5. Adhering to this clause (which is based on the discussion of Section 1‟s Case Study) is clearly the easiest option. BD 37/01 states: “For superstructures for which the fundamental natural frequency of vibration exceeds 5Hz for the unloaded bridge in the vertical direction and 1.” – Appendix B. Once again controlling the fundamental frequency is important.5 Hz for the loaded bridge in the horizontal direction.5 f . The maximum acceleration is then got as umax   2umax (see equations (2.2) where ust is the static deflection under the weight of a pedestrian at the point of maximum deflection.11) for example. and  is the dynamic response factor got again from Figure 5. the vibration serviceability requirement is deemed to be satisfied.Structural Analysis IV Chapter 5 – Structural Dynamics Footbridges As may be gathered from the Case Studies of the Aberfeldy Bridge. Alternatively. the problem is complex.1). This is then compared to a rather simple rule that the maximum acceleration of footbridge decks should not exceed 0. 102 Dr. note:  2  2 f ).5. however some rough guidelines are possible. K is a configuration factor for the type of structure (given in Table 5. C. note from Figure 5.4 the conservative nature of the damping assumed. Table 5. which.35) can be seen to be so based on usual values of damping in structures. from equation (2. C. Table 5. 103 Dr.2: Values of the logarithmic decrement for different bridge types.1: Configuration factors for footbridges. Caprani .Structural Analysis IV Chapter 5 – Structural Dynamics Also. C.Structural Analysis IV Chapter 5 – Structural Dynamics Figure 5.4: Dynamic response factor for footbridges 104 Dr. Caprani . 05 .4 gives   6.21): f   2  182 3  108  3.2835 mm 48  3  108 Table 5. from equations (2.78 m/s 2 2 We compare this to the requirement that: 105 Dr.2835 1.0  6.17  1.93 103  0.1 gives K  1 and Figure 5.19) and (4. Assume the damping to be related to   0. Caprani .9 m and produces an effective pulsating force of 180 N.7 kN pedestrian walking in frequency with the bridge: the pedestrian has a stride of 0. Determine the maximum amplitude of vibration and vertical acceleration caused by a 0.Structural Analysis IV Chapter 5 – Structural Dynamics Design Example A simply-supported footbridge of 18 m span has a total mass of 12. by (5.8 and so.5.2) we have: umax  0. C.8  1. Is this a comfortable bridge for the pedestrian (Figure 5.17 Hz 12600 /18 The static deflection is: 700  183 ust   0.6 tonnes and flexural stiffness of 3×105 kNm2.1)? The natural frequency of the bridge is.93 mm and so the maximum acceleration is: umax   2umax   2  3. 5 f  0. with the amplitude 1.17 Hz frequency. Caprani . soldiers were told to break step when crossing a slender bridge – unfortunately for some.d reveals why. C. historically.93 mm and 3.1. we can see that this pedestrian will feel decidedly uncomfortable and will probably change pace to avoid this frequency of loading.89 m/s 2 And so we deem the bridge acceptable. rather. in conjunction with Section 2.Structural Analysis IV Chapter 5 – Structural Dynamics umax  0. From Figure 5.5 f 0. 106 Dr.78  0. The above discussion. bitter experience. it is more probable that this knowledge did not come from any detailed dynamic analysis.  Any plastic deformation in the joint. It is notable that the materials themselves have very low damping and thus most of the damping observed comes from the joints and so can it depend on:  The materials in contact and their surface preparation. one 107 Dr.3 Damping in Structures The importance of damping should be obvious by this stage. It was alluded to in Section 1 that the exact nature of damping is not really understood but that it has been shown that our assumption of linear viscous damping applies to the majority of structures – a notable exception is soil-structure interaction in which alternative damping models must be assumed.5. Table 5. When the vibrations or DAF is unacceptable it is not generally acceptable to detail joints that will have higher damping than otherwise normal – there are simply too many variables to consider. Caprani . equation (2. Depending on the amount of extra damping needed.47). C.4: Recommended values of damping. a slight increase may significantly reduce the DAF at resonance.3 gives some typical damping values in practice.Structural Analysis IV Chapter 5 – Structural Dynamics 5. Table 5.  The normal force across the interface.  Rubbing or fretting of the joint when it is not tightened. some are:  Tuned mass dampers (TMDs): a relatively small mass is attached to the primary system and is „tuned‟ to vibrate at the same frequency but to oppose the primary system.Structural Analysis IV Chapter 5 – Structural Dynamics could wait for the structure to be built and then measure the damping. In the John Hancock building a concrete block of about 300 tonnes located on the 54th storey sits on a thin film of oil. Caprani . C. the design of a vibration isolation device may be integral to the structure. The devices that may be installed vary. Or. 108 Dr. These are the approaches taken in many modern buildings. This is quite a rudimentary system compared to modern systems which have computer controlled actuators that take input from accelerometers in the building and move the block an appropriate amount. if the extra damping required is significant. particularly in Japan and other earthquake zones. retro-fitting vibration isolation devices as required. The Citicorp building in New York (which is famous for other reasons also) and the John Hancock building in Boston were among the first to use TMDs. When the building sways the inertial effects of the block mean that it moves in the opposite direction to that of the sway and so opposes the motion (relying heavily on a lack of friction). connected at their bases but at opposite sides of the primary system slosh.  Sloshing dampers: A large water tank is used – the sloshing motion opposes the primary system motion due to inertial effects. in a more controlled manner to oppose the primary system motion.  Liquid column dampers: Two columns of liquid. C.a.d).b is probably the best for those structures outside the standard cases of Section 4. Careful thought on reducing the size of the problem to an SDOF system usually enables good approximate analysis.5.c). Machine Loading By avoiding any of the frequencies that the machine operates at. Problematic floors are lightweight with spans of over about 7 m.Structural Analysis IV Chapter 5 – Structural Dynamics 5. vibrations may be minimised. Approximate Frequencies The Bolton Method of Section 4. Caprani . irrespective of magnitude. If the response is still not acceptable vibration isolation devices may need to be considered (Section 5. For normal floors of span/depth ratio less than 25 vibration is not generally a problem. Human loading Most forms of human loading occur at frequencies < 5 Hz (Sections 1 and 5.4 Design Rules of Thumb General The structure should not have any modal frequency close to the frequency of any form of periodic loading. This is based upon the large DAFs that may occur (Section 2.a) and so any structure of natural frequency greater than this should not be subject to undue dynamic excitation. Other methods are: 109 Dr. The addition of either more stiffness or mass will change the frequencies the structure responds to. C.Structural Analysis IV Chapter 5 – Structural Dynamics Structures with concentrated mass: f  1 g 2  Simplified rule for most structures: f  18  where  is the static deflection and g is the acceleration under gravity. Caprani . 110 Dr. Thus.3) This method can be used to estimate the fundamental frequency of MDOF systems. Caprani .5. the fundamental frequency in each direction is given by: Q u  g Q u 2 1 i i i i i 2 i mu g mu i i i i i 2 i (5. and g is the acceleration due to gravity.5. this can be applied to each of the X and Y directions to obtain the estimates of the fundamental sway modes. C. 111 Dr. the first mode is approximated in shape by the static deflection under dead load. for estimating the lowest natural frequency of transverse beam vibration is:  d2y   EI  dx 2  dx 2   1  0 L 2  y dm L 0 2 (5. For a building.5. acting in the direction of motion. Considering the frame of Figure 5. based on energy methods.4) where ui is the static deflection under the dead load of the structure Qi .Structural Analysis IV Chapter 5 – Structural Dynamics Rayleigh Approximation A method developed by Lord Rayleigh (which is always an upper bound). 112 Dr. the fundamental lift and drag modes can be obtained. Caprani .5: Rayleigh approximation for the fundamental sway frequencies of a building. C. The torsional mode can also be approximated by applying the dead load at the appropriate radius of gyration and determining the resulting rotation angle.6.Structural Analysis IV Chapter 5 – Structural Dynamics Figure 5. Figure 5. by applying the dead load in each of the vertical and horizontal directions.6: Rayleigh method for approximating bridge fundamental frequencies. Likewise for a bridge. Figure 5. It provides a reasonable approximate check on the output. Caprani . such as finite-element. C.Structural Analysis IV Chapter 5 – Structural Dynamics This method is particularly useful when considering the results of a detailed analysis. 113 Dr. 5(a) is known to have a static deflection of 32.25 Hz.6.6 Appendix 5.2 mm.7 mm for an unknown mass. 114 Dr. (10%) 3) What time does the first positive peak occur? (10%) 4) What value of damping coefficient is required such that the amplitude after 5 oscillations is 10% of the first peak? (10%) 5) What is the peak force in the spring? (20%) (b) A cantilever riverside boardwalk has been opened to the public as shown in Figure Q.114.1 Past Exam Questions Summer 2005 Question 5 (a) The system shown in Figure Q. 99 kg.5 N. Q. An harmonic oscillation test found the natural frequency of the structure to be 2.5(a) FIG.5(b) Ans. 1) Find the natural frequency of the system. It is proposed to retro-fit braced struts at 5m spacings so that the natural period of vibration will be 9 Hz – given E = 200 kN/mm2 and ignoring buckling effects.and traffic-induced vibrations. however. find the peak displacement when this mass is given an initial velocity of 500 mm/s and an initial displacement of 25 mm.5 mm2.05 s. what area of strut is required? (40%)     k         m FIG. 38. (a) 2.s/m.5(b). it was found that the structure experiences significant human. Caprani . 0. Q. C.Structural Analysis IV Chapter 5 – Structural Dynamics 5.756 Hz. (10%) 2) Given that the mass is 10 kg. (b) 67. 7 mm for an unknown mass. Q5(a) Ans. Q5(b) is loaded with an air conditioning (AC) unit at its tip.      FIG.114. C. 99 kg. (a) 2.2 mm. Q5(a) (b) The beam shown in Fig.91 mm. (2 marks) (ii) Given that the mass is 10 kg.5 N. Q5(a) is known to have a static deflection of 32.s/m. it is found that the maximum steady-state deflection is 20.756 Hz. When the speed of the AC unit is varied. (i) Find the natural frequency of the system. (b) ??. 38. 115 Dr.Structural Analysis IV Sample Paper Semester 1 2006/7 5.05 s. Determine: (i) The damping ratio. (7 marks) Take the following values: • EI = 1×106 kNm2. (2 marks) (iii) At what time does the first positive peak occur? (2 marks) (iv) What damping ratio is required such that the amplitude after 5 oscillations is 10% of the first peak? (2 marks) (v) What is the peak force in the spring? (6 marks)   FIG. 0. find the peak displacement when the mass is given an initial velocity of 500 mm/s and an initial displacement of 25 mm. Caprani . • Mass of the unit is 500 kg. Chapter 5 – Structural Dynamics (a) The single-degree-of-freedom system shown in Fig. The AC unit produces an unbalanced force of 100 kg which varies sinusoidally. (4 marks) (ii) The maximum deflection when the unit‟s speed is 250 rpm. 1%. In your opinion. is the beam cracked or uncracked? Use a single degree-of-freedom (SDOF) system to represent the deflection at the centre of the beam. Its fundamental natural frequency is measured to be 6. • Mass of the unit is 2000 kg. 300 mm wide × 600 mm deep spans 8 m.5 Hz. (a) Cracked. The AC unit produces an unbalanced force of 200 kg which varies sinusoidally. When the speed of the AC unit is  varied.Structural Analysis IV Semester 1 2006/7 5. Chapter 5 – Structural Dynamics (a) A simply-supported reinforced concrete beam. Caprani . Take the following values: • EI = 40×103 kNm2. it is found that the maximum steady-state amplitude of vibration is 34.6 mm.  (5 marks) (10 marks) The maximum deflection when the unit‟s speed is 100 rpm. C. Q5(a)      FIG.1 mm. 116 Dr. (10 marks) (b) The beam shown in Fig. 41. FIG. Q5(b) is loaded with an air conditioning (AC) unit at its tip. Determine: (i) (ii) The damping ratio. Q5(b) Ans. Take the density of reinforced concrete to be 24 kN/m3 and E = 30 kN/mm2. • Ignore the mass of the beam. (b) 5. Assume that 8/15 of the total mass of the beam contributes to the SDOF model. 67 mm. 0. of the damped behaviour compared to the undamped behaviour? (10 marks) (c) A machine is placed on member BC which has an unbalanced force of 500 kg which varies sinusoidally.        FIG. (b) Ratio: 28.4%. (8 marks) (b) The frame is found to have 5% damping.93 Hz.  M = 20 tonnes. determine: (i) The natural frequency and period in free vibration. 20cos[24. Using appropriate approximations. (ii) the speed of the machine at resonance. (7 marks) Note: Take the following values:  EI = 20×103 kNm2.  Consider BC as infinitely rigid.   117 Dr. C. (iii) the displacement at resonance.01 mm. 4. change: 71. what is the percentage change in deflection.(a) 3. Neglecting the mass of the machine. Caprani . (ii) An expression for the displacement at time t if member BC is displaced 20 mm and suddenly released at time t = 1 sec. Q5. 4 cycles after the frame is released. 236 rpm.6%. Q5 Ans. determine: (i) the maximum displacement when the unit‟s speed is 150 rpm.Structural Analysis IV Semester 1 2007/8 QUESTION 5 Chapter 5 – Structural Dynamics (a) For the frame shown in Fig. using a single-degree-of-freedom model.72(t-1)]. t>1.254 s. (c) 0. (v) the displacement at resonance. determine: (i) the natural frequency and period in free vibration. 0.  M = 5 tonnes. 2%.2 mm.2 rpm. (ii) the damping. Using a single-degree-of-freedom model for vibrations in the vertical direction.008 mm.9 Hz. Q5 supports a scoreboard at a sports centre. and neglecting the mass of the truss members.30 mm. 0. C. Caprani . 4. the amplitude was 5. Q5 Ans. (iii) the maximum displacement when the unit‟s speed is 1500 rpm. given that a test showed 5 cycles after a 10 mm initial displacement was imposed. (iv) the speed of the machine at resonance. 293.          FIG. (25 marks) Note: Take the following values:  For all truss members: EA  20 103 kN .205 s. 5.  118 Dr.  Ignore the stiffness and mass of member EF.Structural Analysis IV Chapter 5 – Structural Dynamics Semester 1 2008/9 QUESTION 5 The structure shown in Fig. The claxton (of total mass M) which sounds the end of playing periods includes a motor which has an unbalanced mass of 100 kg which varies sinusoidally when sounded. Determine the damping of the system. calculate the natural frequency in lateral vibration and its period. At resonance. Find the force required to deflect the frame 20 mm laterally. 48 kN. Assuming the beam to be infinitely rigid.7 103 kNm2 . the amplitude of vibration is found to be 10 mm.07 mm. 6 m wide single-bay single-storey frame is rigidly jointed with a beam of mass 2. 0..80 of the resonant frequency.51 Hz.Structural Analysis IV Semester 1 2009/10 QUESTION 5 Chapter 5 – Structural Dynamics (a) A 3 m high. C. (10 marks) (b) A spring-mass-damper SDOF system is subject to a harmonically varying force. Caprani .1. (15 marks) Ans. 5.000 kg and columns of negligible mass and stiffness of EI  2. and at 0. 119 Dr. the amplitude is found to be 5. Structural Analysis IV Semester 1 2010/11 QUESTION 5 Chapter 5 – Structural Dynamics (a) For the shear frame shown in Fig. Q5(a), ignoring the mass of the columns: (i) How many modes will this structure have? (ii) Sketch the mode shapes; (iii) Indicate the order of the natural frequencies associated with each mode shape (i.e. lowest to highest). (10 marks) (b) For the frame shown in Fig. Q5(b), using a single-degree-of-freedom model, determine the natural frequency and period in free vibration given that EI = 27×103 kNm2 and M = 24 tonnes. If a machine is placed on member BC which has an unbalanced force of 500 kg varying sinusoidally, neglecting the mass of the machine, determine: (i) the maximum displacement when the unit‟s speed is 360 rpm; (ii) the speed of the machine at resonance; (iii) the displacement at resonance. (15 marks)       FIG. Q5(a)         FIG. Q5(b) Ans. 0.34 mm, 426.6 rpm, 1.02 mm. 120 Dr. C. Caprani Structural Analysis IV Chapter 5 – Structural Dynamics 5.6.2 References The following books/articles were referred to in the writing of these notes; particularly Clough & Penzien (1993), Smith (1988) and Bolton (1978) - these should be referred to first for more information. There is also a lot of information and software available online; the software can especially help intuitive understanding. The class notes of Mr. R. Mahony (D.I.T.) and Dr. P. Fanning (U.C.D.) were also used.  Archbold, P., (2002), “Modal Analysis of a GRP Cable-Stayed Bridge”, Proceedings of the First Symposium of Bridge Engineering Research In Ireland, Eds. C. McNally & S. Brady, University College Dublin.  Beards, C.F., (1983), Structural Vibration Analysis: modelling, analysis and damping of vibrating structures, Ellis Horwood, Chichester, England.  Bhatt, P., (1999), Structures, Longman, Harlow, England.  Bolton, A., (1978), “Natural frequencies of structures for designers”, The Structural Engineer, Vol. 56A, No. 9, pp. 245-253; Discussion: Vol. 57A, No. 6, p.202, 1979.  Bolton, A., (1969), “The natural frequencies of continuous beams”, The Structural Engineer, Vol. 47, No. 6, pp.233-240.  Case, J., Chilver, A.H. and Ross, C.T.F., (1999), Strength of Materials and Structures, 4th edn., Arnold, London.  Chopra, A.K., (2007), Dynamics of Structures – Theory and Applications to Earthquake Engineering, 3rd edn., Pearson-Prentice Hall, New Jersey.  Clough, R.W. and Penzien, J., (1993), Dynamics of Structures, 2nd edn., McGrawHill, New York.  Cobb, F. (2004), Structural Engineer’s Pocket Book, Elsevier, Oxford.  Craig, R.R., (1981), Structural Dynamics – An introduction to computer methods, Wiley, New York. 121 Dr. C. Caprani Structural Analysis IV Chapter 5 – Structural Dynamics  Ghali, A. and Neville, A.M., (1997), Structural Analysis – A unified classical and matrix approach, 4th edn., E&FN Spon, London.  Irvine, M., (1986), Structural Dynamics for the Practising Engineer, Allen & Unwin, London.  Kreyszig, E., (1993), Advanced Engineering Mathematics, 7th edn., Wiley.  Paz, M. and Leigh, W., (2004), Structural Dynamics – Theory and Computation, 5th edn., Springer, New York.  Smith, J.W., (1988), Vibration of Structures – Applications in civil engineering design, Chapman and Hall, London. 122 Dr. C. Caprani 6. we regularly wish to express it in one of the following forms: u  t   C cos t    u  t   C cos t    (5.5.11) If we consider that A.5.7) Where C  A2  B 2 tan   tan   (5.5.5. then we can draw the following: 123 Dr.5.5. re-write equation (5. B and C represent a right-angled triangle with angles  and  .8) (5.Structural Analysis IV Chapter 5 – Structural Dynamics 5.10) A B B A To arrive at this result.5.5) However.3 Amplitude Solution to Equation of Motion The solution to the equation of motion is found to be in the form: u  t   A cos t  B sin t (5.5) as: B A  u  t   C  cos t  sin t  C C  (5. C. Caprani .9) (5.6) (5.5. 18) (5.5. C.14) (5.16) (5.15) And using the well-known trigonometric identities: sin  X  Y   sin X cos Y  cos X sin Y cos  X  Y   cos X cos Y  sin X sin Y (5.5.5.5.5.5.11) gives two relationships: u  t   C sin  cos t  cos sin t  u  t   C cos  cos t  sin  sin t  (5.5.13) Introducing these into equation (5. the last of which is the one we adopt: u  t   C sin t    (5.19) u  t   C cos t    124 Dr.5.12) (5.Structural Analysis IV Chapter 5 – Structural Dynamics Thus: sin   cos   cos   sin   A C B C (5.17) Gives the two possible representations.5. Caprani . 4 Solutions to Differential Equations The Homogenous Equation To find the solution of: d2y  k2y  0 2 dx (5.5. dx d2y   2e x 2 dx Substituting this into (5. Caprani .20) we try y  e x (note that this k has nothing to do with stiffness but is the conventional mathematical notation for this problem). C. Thus we have: dy   e x .20) gives:  2 e x  k 2 e  x  0 And so we get the characteristic equation by dividing out e x : 2  k2  0 From which:    k 2 Or. 125 Dr.Structural Analysis IV Chapter 5 – Structural Dynamics 5.6.5. 2  ik Where i  1 . Caprani .5.5. Since these are both solutions.23) 126 Dr.Structural Analysis IV Chapter 5 – Structural Dynamics 1  ik . Introducing Euler‟s equations: eikx  cos kx  i sin kx e ikx  cos kx  i sin kx into (5.21) In which A1 and A2 are constants to be determined from the initial conditions of the problem. C.5.21) gives us: (5. they are both valid and the expression for y becomes: y  Aeikx  A2e ikx 1 (5.5.22) y  A1  cos kx  i sin kx   A2  cos kx  i sin kx  Collecting terms: y   A1  A2  cos kx   iA1  iA2  sin kx Since the coefficients of the trigonometric functions are constants we can just write: y  A cos kx  B sin kx (5. 2.26) For the particular solution we try the following: uP  t   C sin t  D cos t (5.  The particular solution ( uP  t  ): particular to the function on the right hand side of equation (5. The final solution is the sum of the complimentary and particular solutions: u  t   uC  t   uP  t  (5. Caprani .5.2.27) Then we have: uP  t   C cos t  D sin t 127 (5. which we must now find.10) and (5.5.24) We divide by m and introduce equations (5.5.5.28) Dr.5.5.23)). which we already have (equation (5.47) (repeated here for convenience): mu(t )  cu(t )  ku(t )  F0 sin t (5.Structural Analysis IV Chapter 5 – Structural Dynamics The Non-homogenous Equation Starting with equation (5.5.24). C. recall that the solution to non-homogenous differential equations is made up of two parts:  The complimentary solution ( uC  t  ): this is the solution to the corresponding homogenous equation.12) to get: u (t )  2u (t )   2u (t )  F0 sin t m (5.2.25) At this point. 27).29) into equation (5.5. C.5.29) Substituting equations (5.32) (5.5.30) Collecting sine and cosine terms:  2   2  C  2D  sin t   F   2C      D  cos t  0 sin t   m 2 2 (5.5. Caprani . divide both sides by  2 : 128 Dr.5. the sine and cosine terms on both sides of the equation must be equal.28) and (5. (5. Thus:  2  2  C  2D  F0 m (5.31) For this to be valid for all t.5.5.5.5.25) gives:   2C sin t   2 D cos t    2  C cos t  D sin t   2 C sin t  D cos t   F0 sin t m (5.Structural Analysis IV Chapter 5 – Structural Dynamics And uP  t   2C sin t  2 D cos t (5.33) 2C   2  2  D  0 Next. 36) (5.5.37). equation (5. C.36) gives:  1   2 2  F  2  D  0  2 k     (5. Caprani .     .5.51). we have: 1    D C 2 2 (5.40) 129 Dr.5.5.9) to get: 1    C  2 D  2 F0 k (5.5.5. and k   2 m from equation (5.39) To get:  1   2 2   2 2  F  D  0 2 k     (5.2.5.38) And using this in equation (5.2.Structural Analysis IV Chapter 5 – Structural Dynamics  F  2  1  2  C  2 D  20    m    2  2 C  1  2  D  0      (5.34) (5.5.35) Introduce the frequency ratio.5.37) 2 C  1   2  D  0 From equation (5. finally: 1    F C 0 2 k 1   2    2 2 2 (5.44) (5. finally: F0 2 2 k 1   2    2 2 D (5. we have: 1     F C  2 2   2 2  k 1   2    2     0 2 (5. C.5.41) Now using this with equation (5.5.45) To express the solution as: uP (t )   sin  t    (5.5.5.41): 130 Dr.5.5.43) and (5.38). from equations (5.43) Again we use the cosine addition rule:   C 2  D2 tan    D C (5.5.5.5.42) To get.46) So we have.5. (5.44).Structural Analysis IV Chapter 5 – Structural Dynamics And rearrange to get. Caprani . 43) and (5.51) 131 Dr.5.41) again to get: F0 2 2 k 1   2    2 2 tan    1    F0 2 k 1   2    2 2 2 (5.5.49) To obtain the phase angle.5. Caprani . we use equation (5.5.Structural Analysis IV Chapter 5 – Structural Dynamics  F0 k   1   2     2     2 2 2 2  1   2    2    1   2    2       2 2 (5.5.5.47) This simplifies to: F  0 k 1      2  1      2     2 2 2 2 2 2 2 (5.45) with equations (5. C.50) Immediately we see that several terms (and the minus signs) cancel to give: tan   2 1  2 (5.5.48) And finally we have the amplitude of displacement:  F0  2 2 2 2  1      2   k  1 (5.5. 51).5.Structural Analysis IV Chapter 5 – Structural Dynamics Thus we have the final particular solution of equation (5.5. u0 and u0 we differentiate equation (5. the total solution is the sum of the particular and complimentary solutions.5.53) to get: u  t   et d B  A  cos d t   Ad  B  sin d t    cos  t    (5. C.5. we have from equations (5.5.5.54): u  0   u0  A   sin  (5.54)   Now at t  0 .35) since we have redefined the amplitude and phase in terms of the forcing function.55) And: u  0  u0  d B  A    cos (5.5.5.53) Notice here that we used equation (5.55) gives: A  u0   sin  132 (5.49) and (5.52) (5. which for us now becomes: u  t   uC  t   uP  t  u  t   et  A cos d t  B sin d t    sin  t    (5.5.57) Dr.2.5. To determine the unknown constants from the initial parameters.53) and (5. As referred to previously.5.5.5.56) Solving for A first from equation (5.46) in conjunction with equations (5. Caprani . 58) Multiplying out and rearranging gives: u0  d B  u0     cos   sin  (5. C. Caprani .5.Structural Analysis IV Chapter 5 – Structural Dynamics And introducing this into equation (5.5.59) From which we get: u0  u0     cos   sin   B d (5.  The particular solution ( uP  t  ): represents the steady state of the system which persists as long as the harmonic force is applied. 133 Dr. as again may be seen from equation (5.5.5.53).5. Remember that:  The complimentary solution ( uC  t  ): represents the transient state of the system which dampens out after a period of time. in addition to the exponentially reducing term in equation (5.56) gives: u0  d B   u0   sin    cos (5.53).60) And now we have completely defined the time history of the problem in terms of its initial parameters. as may be realized when it is seen that it is only the complementary response that is affected by the initial state ( u0 and u0 ) of the system.5. period and frequency d   1   2 Td  2 d . Caprani .6. C. Coefficient of damping Circular frequency Damping ratio Critical value of damping mu(t )  cu(t )  ku(t )  F (t ) u(t )  2u(t )   2u(t )  0 f  1  1   T 2 2 k m 2  c m 2   k m c ccr ccr  2m  2 km u(t )   cos t    General solution for free-undamped vibration u u    u   0  . d   2 0 2 General solution for free-damped vibrations tan   u0  u0 u0d 134 Dr. period.5 Important Formulae SDOF Systems Fundamental equation of motion Equation of motion for free vibration Relationship between frequency. fd  d 2 u(t )   et cos d t     u  u0    u  0  . stiffness and mass: Fundamental frequency for an SDOF system. circular frequency. tan   0 u0   2 0 2 Damped circular frequency.Structural Analysis IV Chapter 5 – Structural Dynamics 5. Caprani .Structural Analysis IV Chapter 5 – Structural Dynamics Logarithmic decrement of damping Half-amplitude method   ln un   2m un  m d  un  p 0.11 when unm  0. C.  k  response and frequency ratio  2  tan   1  2  2 2 DAF  D  1   2    2     1 2 Dynamic amplification factor (DAF) 135 Dr.5un m u    n1  un  un  p Amplitude after p-cycles Equation of motion for forced response (sinusoidal) mu(t )  cu(t )  ku(t )  F0 sin t u p  t    sin  t    1 2 General solution for forced-damped vibration   F0 1   2 2   2 2  . C.Structural Analysis IV Chapter 5 – Structural Dynamics MDOF Systems Fundamental equation of motion Equation of motion for undamped-free vibration Mu + Cu + Ku = F Mu + Ku = 0 General solution and derivates u = a sin t    for free-undamped vibration Frequency equation General solution for 2DOF system Determinant of 2DOF system from Cramer‟s rule Composite matrix Amplitude equation u =  2a sin t      2u K   2M  a = 0    m1 0  0   u1   k1  k2    m2  u2   k2  k2   u1  0      k2  u2  0 K   2M =  k2  k1    2m1  k2   2m2   k22  0    E  K   2M    Ea = 0 136 Dr. Caprani . t  t 2  p  x. t   0 Boundary conditions for a simply x supported beam Frequencies of a simply supported beam Mode shape or mode n: (A1 is normally unity) Cantilever beam boundary conditions Frequency equation for a cantilever sin( x)  sinh( x)     sin( L)  sinh( L)   n  x   A1    cos( L)  cosh( L)    cosh( x)  cos( x)     f  1 2 KE ME v  L. t   0 and  n x    L  v  0. Caprani .Structural Analysis IV Chapter 5 – Structural Dynamics Continuous Structures Equation of motion Assumed solution for freeundamped vibrations General solution EI  4v  x. t   0 x 2 EI m n  x   A1 sin  v  0. t  x 4 m  2 v  x. t   0 and EI 3  L. t   0 x x cos( L)cosh( L)  1  0 Cantilever mode shapes Bolton method general equation 137 Dr. t   0 and EI 2  0. t     x  Y  t    x   A1 sin  x   A2 cos  x   A3 sinh  x   A4 cosh  x   2v v  0. t  v  x. C. t   0 and EI  n  n     L  2  2v  L. t   0 x  2v  3v EI 2  L. Structural Analysis IV Chapter 5 – Structural Dynamics Practical Design a0   0. Caprani .9  2 f I M Peak acceleration under foot-loading I  70 Ns M  40% mass per unit area Maximum dynamic deflection Maximum vertical acceleration umax  ust K umax   2umax BD37/01 requirement for vertical acceleration 0. C.5 f 138 Dr. 6. C. Fill this out as you progress through the notes.Structural Analysis IV Chapter 5 – Structural Dynamics 5.6 Glossary Structural dynamics introduces many new terms and concepts so it‟s beneficial to keep track of them in one place. Caprani . Symbol Name Units 139 Dr.